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Popular Trigonometry >

cosh(4x)=16sinh(x)+1

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Solution

cosh(4x)=16sinh(x)+1

Solution

x=0,x=ln(2.41421…)
+1
Degrees
x=0∘,x=50.49898…∘
Solution steps
cosh(4x)=16sinh(x)+1
Rewrite using trig identities
cosh(4x)=16sinh(x)+1
Use the Hyperbolic identity: sinh(x)=2ex−e−x​cosh(4x)=16⋅2ex−e−x​+1
Use the Hyperbolic identity: cosh(x)=2ex+e−x​2e4x+e−4x​=16⋅2ex−e−x​+1
2e4x+e−4x​=16⋅2ex−e−x​+1
2e4x+e−4x​=16⋅2ex−e−x​+1:x=0,x=ln(2.41421…)
2e4x+e−4x​=16⋅2ex−e−x​+1
Multiply both sides by 22e4x+e−4x​⋅2=16⋅2ex−e−x​⋅2+1⋅2
Simplifye4x+e−4x=16(ex−e−x)+2
Apply exponent rules
e4x+e−4x=16(ex−e−x)+2
Apply exponent rule: abc=(ab)ce4x=(ex)4,e−4x=(ex)−4,e−x=(ex)−1(ex)4+(ex)−4=16(ex−(ex)−1)+2
(ex)4+(ex)−4=16(ex−(ex)−1)+2
Rewrite the equation with ex=u(u)4+(u)−4=16(u−(u)−1)+2
Solve u4+u−4=16(u−u−1)+2:u=−1,u=1,u≈−0.41421…,u≈2.41421…
u4+u−4=16(u−u−1)+2
Refineu4+u41​=16(u−u1​)+2
Multiply both sides by u4
u4+u41​=16(u−u1​)+2
Multiply both sides by u4u4u4+u41​u4=16(u−u1​)u4+2u4
Simplify
u4u4+u41​u4=16(u−u1​)u4+2u4
Simplify u4u4:u8
u4u4
Apply exponent rule: ab⋅ac=ab+cu4u4=u4+4=u4+4
Add the numbers: 4+4=8=u8
Simplify u41​u4:1
u41​u4
Multiply fractions: a⋅cb​=ca⋅b​=u41⋅u4​
Cancel the common factor: u4=1
u8+1=16(u−u1​)u4+2u4
u8+1=16(u−u1​)u4+2u4
u8+1=16(u−u1​)u4+2u4
Expand 16(u−u1​)u4+2u4:16u5−16u3+2u4
16(u−u1​)u4+2u4
=16u4(u−u1​)+2u4
Expand 16u4(u−u1​):16u5−16u3
16u4(u−u1​)
Apply the distributive law: a(b−c)=ab−aca=16u4,b=u,c=u1​=16u4u−16u4u1​
=16u4u−16⋅u1​u4
Simplify 16u4u−16⋅u1​u4:16u5−16u3
16u4u−16⋅u1​u4
16u4u=16u5
16u4u
Apply exponent rule: ab⋅ac=ab+cu4u=u4+1=16u4+1
Add the numbers: 4+1=5=16u5
16⋅u1​u4=16u3
16⋅u1​u4
Multiply fractions: a⋅cb​=ca⋅b​=u1⋅16u4​
Multiply the numbers: 1⋅16=16=u16u4​
Cancel the common factor: u=16u3
=16u5−16u3
=16u5−16u3
=16u5−16u3+2u4
u8+1=16u5−16u3+2u4
Solve u8+1=16u5−16u3+2u4:u=−1,u=1,u≈−0.41421…,u≈2.41421…
u8+1=16u5−16u3+2u4
Move 2u4to the left side
u8+1=16u5−16u3+2u4
Subtract 2u4 from both sidesu8+1−2u4=16u5−16u3+2u4−2u4
Simplifyu8+1−2u4=16u5−16u3
u8+1−2u4=16u5−16u3
Move 16u3to the left side
u8+1−2u4=16u5−16u3
Add 16u3 to both sidesu8+1−2u4+16u3=16u5−16u3+16u3
Simplifyu8+1−2u4+16u3=16u5
u8+1−2u4+16u3=16u5
Move 16u5to the left side
u8+1−2u4+16u3=16u5
Subtract 16u5 from both sidesu8+1−2u4+16u3−16u5=16u5−16u5
Simplifyu8+1−2u4+16u3−16u5=0
u8+1−2u4+16u3−16u5=0
Write in the standard form an​xn+…+a1​x+a0​=0u8−16u5−2u4+16u3+1=0
Factor u8−16u5−2u4+16u3+1:(u+1)(u−1)(u6+u4−16u3−u2−1)
u8−16u5−2u4+16u3+1
Use the rational root theorem
a0​=1,an​=1
The dividers of a0​:1,The dividers of an​:1
Therefore, check the following rational numbers:±11​
−11​ is a root of the expression, so factor out u+1
=(u+1)u+1u8−16u5−2u4+16u3+1​
u+1u8−16u5−2u4+16u3+1​=u7−u6+u5−17u4+15u3+u2−u+1
u+1u8−16u5−2u4+16u3+1​
Divide u+1u8−16u5−2u4+16u3+1​:u+1u8−16u5−2u4+16u3+1​=u7+u+1−u7−16u5−2u4+16u3+1​
Divide the leading coefficients of the numerator u8−16u5−2u4+16u3+1
and the divisor u+1:uu8​=u7
Quotient=u7
Multiply u+1 by u7:u8+u7Subtract u8+u7 from u8−16u5−2u4+16u3+1 to get new remainderRemainder=−u7−16u5−2u4+16u3+1
Thereforeu+1u8−16u5−2u4+16u3+1​=u7+u+1−u7−16u5−2u4+16u3+1​
=u7+u+1−u7−16u5−2u4+16u3+1​
Divide u+1−u7−16u5−2u4+16u3+1​:u+1−u7−16u5−2u4+16u3+1​=−u6+u+1u6−16u5−2u4+16u3+1​
Divide the leading coefficients of the numerator −u7−16u5−2u4+16u3+1
and the divisor u+1:u−u7​=−u6
Quotient=−u6
Multiply u+1 by −u6:−u7−u6Subtract −u7−u6 from −u7−16u5−2u4+16u3+1 to get new remainderRemainder=u6−16u5−2u4+16u3+1
Thereforeu+1−u7−16u5−2u4+16u3+1​=−u6+u+1u6−16u5−2u4+16u3+1​
=u7−u6+u+1u6−16u5−2u4+16u3+1​
Divide u+1u6−16u5−2u4+16u3+1​:u+1u6−16u5−2u4+16u3+1​=u5+u+1−17u5−2u4+16u3+1​
Divide the leading coefficients of the numerator u6−16u5−2u4+16u3+1
and the divisor u+1:uu6​=u5
Quotient=u5
Multiply u+1 by u5:u6+u5Subtract u6+u5 from u6−16u5−2u4+16u3+1 to get new remainderRemainder=−17u5−2u4+16u3+1
Thereforeu+1u6−16u5−2u4+16u3+1​=u5+u+1−17u5−2u4+16u3+1​
=u7−u6+u5+u+1−17u5−2u4+16u3+1​
Divide u+1−17u5−2u4+16u3+1​:u+1−17u5−2u4+16u3+1​=−17u4+u+115u4+16u3+1​
Divide the leading coefficients of the numerator −17u5−2u4+16u3+1
and the divisor u+1:u−17u5​=−17u4
Quotient=−17u4
Multiply u+1 by −17u4:−17u5−17u4Subtract −17u5−17u4 from −17u5−2u4+16u3+1 to get new remainderRemainder=15u4+16u3+1
Thereforeu+1−17u5−2u4+16u3+1​=−17u4+u+115u4+16u3+1​
=u7−u6+u5−17u4+u+115u4+16u3+1​
Divide u+115u4+16u3+1​:u+115u4+16u3+1​=15u3+u+1u3+1​
Divide the leading coefficients of the numerator 15u4+16u3+1
and the divisor u+1:u15u4​=15u3
Quotient=15u3
Multiply u+1 by 15u3:15u4+15u3Subtract 15u4+15u3 from 15u4+16u3+1 to get new remainderRemainder=u3+1
Thereforeu+115u4+16u3+1​=15u3+u+1u3+1​
=u7−u6+u5−17u4+15u3+u+1u3+1​
Divide u+1u3+1​:u+1u3+1​=u2+u+1−u2+1​
Divide the leading coefficients of the numerator u3+1
and the divisor u+1:uu3​=u2
Quotient=u2
Multiply u+1 by u2:u3+u2Subtract u3+u2 from u3+1 to get new remainderRemainder=−u2+1
Thereforeu+1u3+1​=u2+u+1−u2+1​
=u7−u6+u5−17u4+15u3+u2+u+1−u2+1​
Divide u+1−u2+1​:u+1−u2+1​=−u+u+1u+1​
Divide the leading coefficients of the numerator −u2+1
and the divisor u+1:u−u2​=−u
Quotient=−u
Multiply u+1 by −u:−u2−uSubtract −u2−u from −u2+1 to get new remainderRemainder=u+1
Thereforeu+1−u2+1​=−u+u+1u+1​
=u7−u6+u5−17u4+15u3+u2−u+u+1u+1​
Divide u+1u+1​:u+1u+1​=1
Divide the leading coefficients of the numerator u+1
and the divisor u+1:uu​=1
Quotient=1
Multiply u+1 by 1:u+1Subtract u+1 from u+1 to get new remainderRemainder=0
Thereforeu+1u+1​=1
=u7−u6+u5−17u4+15u3+u2−u+1
=u7−u6+u5−17u4+15u3+u2−u+1
Factor u7−u6+u5−17u4+15u3+u2−u+1:(u−1)(u6+u4−16u3−u2−1)
u7−u6+u5−17u4+15u3+u2−u+1
Use the rational root theorem
a0​=1,an​=1
The dividers of a0​:1,The dividers of an​:1
Therefore, check the following rational numbers:±11​
11​ is a root of the expression, so factor out u−1
=(u−1)u−1u7−u6+u5−17u4+15u3+u2−u+1​
u−1u7−u6+u5−17u4+15u3+u2−u+1​=u6+u4−16u3−u2−1
u−1u7−u6+u5−17u4+15u3+u2−u+1​
Divide u−1u7−u6+u5−17u4+15u3+u2−u+1​:u−1u7−u6+u5−17u4+15u3+u2−u+1​=u6+u−1u5−17u4+15u3+u2−u+1​
Divide the leading coefficients of the numerator u7−u6+u5−17u4+15u3+u2−u+1
and the divisor u−1:uu7​=u6
Quotient=u6
Multiply u−1 by u6:u7−u6Subtract u7−u6 from u7−u6+u5−17u4+15u3+u2−u+1 to get new remainderRemainder=u5−17u4+15u3+u2−u+1
Thereforeu−1u7−u6+u5−17u4+15u3+u2−u+1​=u6+u−1u5−17u4+15u3+u2−u+1​
=u6+u−1u5−17u4+15u3+u2−u+1​
Divide u−1u5−17u4+15u3+u2−u+1​:u−1u5−17u4+15u3+u2−u+1​=u4+u−1−16u4+15u3+u2−u+1​
Divide the leading coefficients of the numerator u5−17u4+15u3+u2−u+1
and the divisor u−1:uu5​=u4
Quotient=u4
Multiply u−1 by u4:u5−u4Subtract u5−u4 from u5−17u4+15u3+u2−u+1 to get new remainderRemainder=−16u4+15u3+u2−u+1
Thereforeu−1u5−17u4+15u3+u2−u+1​=u4+u−1−16u4+15u3+u2−u+1​
=u6+u4+u−1−16u4+15u3+u2−u+1​
Divide u−1−16u4+15u3+u2−u+1​:u−1−16u4+15u3+u2−u+1​=−16u3+u−1−u3+u2−u+1​
Divide the leading coefficients of the numerator −16u4+15u3+u2−u+1
and the divisor u−1:u−16u4​=−16u3
Quotient=−16u3
Multiply u−1 by −16u3:−16u4+16u3Subtract −16u4+16u3 from −16u4+15u3+u2−u+1 to get new remainderRemainder=−u3+u2−u+1
Thereforeu−1−16u4+15u3+u2−u+1​=−16u3+u−1−u3+u2−u+1​
=u6+u4−16u3+u−1−u3+u2−u+1​
Divide u−1−u3+u2−u+1​:u−1−u3+u2−u+1​=−u2+u−1−u+1​
Divide the leading coefficients of the numerator −u3+u2−u+1
and the divisor u−1:u−u3​=−u2
Quotient=−u2
Multiply u−1 by −u2:−u3+u2Subtract −u3+u2 from −u3+u2−u+1 to get new remainderRemainder=−u+1
Thereforeu−1−u3+u2−u+1​=−u2+u−1−u+1​
=u6+u4−16u3−u2+u−1−u+1​
Divide u−1−u+1​:u−1−u+1​=−1
Divide the leading coefficients of the numerator −u+1
and the divisor u−1:u−u​=−1
Quotient=−1
Multiply u−1 by −1:−u+1Subtract −u+1 from −u+1 to get new remainderRemainder=0
Thereforeu−1−u+1​=−1
=u6+u4−16u3−u2−1
=u6+u4−16u3−u2−1
=(u−1)(u6+u4−16u3−u2−1)
=(u+1)(u−1)(u6+u4−16u3−u2−1)
(u+1)(u−1)(u6+u4−16u3−u2−1)=0
Using the Zero Factor Principle: If ab=0then a=0or b=0u+1=0oru−1=0oru6+u4−16u3−u2−1=0
Solve u+1=0:u=−1
u+1=0
Move 1to the right side
u+1=0
Subtract 1 from both sidesu+1−1=0−1
Simplifyu=−1
u=−1
Solve u−1=0:u=1
u−1=0
Move 1to the right side
u−1=0
Add 1 to both sidesu−1+1=0+1
Simplifyu=1
u=1
Solve u6+u4−16u3−u2−1=0:u≈−0.41421…,u≈2.41421…
u6+u4−16u3−u2−1=0
Find one solution for u6+u4−16u3−u2−1=0 using Newton-Raphson:u≈−0.41421…
u6+u4−16u3−u2−1=0
Newton-Raphson Approximation Definition
f(u)=u6+u4−16u3−u2−1
Find f′(u):6u5+4u3−48u2−2u
dud​(u6+u4−16u3−u2−1)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(u6)+dud​(u4)−dud​(16u3)−dud​(u2)−dud​(1)
dud​(u6)=6u5
dud​(u6)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=6u6−1
Simplify=6u5
dud​(u4)=4u3
dud​(u4)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=4u4−1
Simplify=4u3
dud​(16u3)=48u2
dud​(16u3)
Take the constant out: (a⋅f)′=a⋅f′=16dud​(u3)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=16⋅3u3−1
Simplify=48u2
dud​(u2)=2u
dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=2u2−1
Simplify=2u
dud​(1)=0
dud​(1)
Derivative of a constant: dxd​(a)=0=0
=6u5+4u3−48u2−2u−0
Simplify=6u5+4u3−48u2−2u
Let u0​=1Compute un+1​ until Δun+1​<0.000001
u1​=0.6:Δu1​=0.4
f(u0​)=16+14−16⋅13−12−1=−16f′(u0​)=6⋅15+4⋅13−48⋅12−2⋅1=−40u1​=0.6
Δu1​=∣0.6−1∣=0.4Δu1​=0.4
u2​=0.32945…:Δu2​=0.27054…
f(u1​)=0.66+0.64−16⋅0.63−0.62−1=−4.639744f′(u1​)=6⋅0.65+4⋅0.63−48⋅0.62−2⋅0.6=−17.14944u2​=0.32945…
Δu2​=∣0.32945…−0.6∣=0.27054…Δu2​=0.27054…
u3​=0.03701…:Δu3​=0.29243…
f(u2​)=0.32945…6+0.32945…4−16⋅0.32945…3−0.32945…2−1=−1.66761…f′(u2​)=6⋅0.32945…5+4⋅0.32945…3−48⋅0.32945…2−2⋅0.32945…=−5.70244…u3​=0.03701…
Δu3​=∣0.03701…−0.32945…∣=0.29243…Δu3​=0.29243…
u4​=−7.14264…:Δu4​=7.17966…
f(u3​)=0.03701…6+0.03701…4−16⋅0.03701…3−0.03701…2−1=−1.00217…f′(u3​)=6⋅0.03701…5+4⋅0.03701…3−48⋅0.03701…2−2⋅0.03701…=−0.13958…u4​=−7.14264…
Δu4​=∣−7.14264…−0.03701…∣=7.17966…Δu4​=7.17966…
u5​=−5.91974…:Δu5​=1.22290…
f(u4​)=(−7.14264…)6+(−7.14264…)4−16(−7.14264…)3−(−7.14264…)2−1=141168.16963…f′(u4​)=6(−7.14264…)5+4(−7.14264…)3−48(−7.14264…)2−2(−7.14264…)=−115436.50403…u5​=−5.91974…
Δu5​=∣−5.91974…−(−7.14264…)∣=1.22290…Δu5​=1.22290…
u6​=−4.88878…:Δu6​=1.03095…
f(u5​)=(−5.91974…)6+(−5.91974…)4−16(−5.91974…)3−(−5.91974…)2−1=47545.59081…f′(u5​)=6(−5.91974…)5+4(−5.91974…)3−48(−5.91974…)2−2(−5.91974…)=−46117.92631…u6​=−4.88878…
Δu6​=∣−4.88878…−(−5.91974…)∣=1.03095…Δu6​=1.03095…
u7​=−4.01362…:Δu7​=0.87515…
f(u6​)=(−4.88878…)6+(−4.88878…)4−16(−4.88878…)3−(−4.88878…)2−1=16068.08422…f′(u6​)=6(−4.88878…)5+4(−4.88878…)3−48(−4.88878…)2−2(−4.88878…)=−18360.23113…u7​=−4.01362…
Δu7​=∣−4.01362…−(−4.88878…)∣=0.87515…Δu7​=0.87515…
u8​=−3.26329…:Δu8​=0.75033…
f(u7​)=(−4.01362…)6+(−4.01362…)4−16(−4.01362…)3−(−4.01362…)2−1=5457.33921…f′(u7​)=6(−4.01362…)5+4(−4.01362…)3−48(−4.01362…)2−2(−4.01362…)=−7273.21083…u8​=−3.26329…
Δu8​=∣−3.26329…−(−4.01362…)∣=0.75033…Δu8​=0.75033…
u9​=−2.61197…:Δu9​=0.65132…
f(u8​)=(−3.26329…)6+(−3.26329…)4−16(−3.26329…)3−(−3.26329…)2−1=1865.40737…f′(u8​)=6(−3.26329…)5+4(−3.26329…)3−48(−3.26329…)2−2(−3.26329…)=−2864.03467…u9​=−2.61197…
Δu9​=∣−2.61197…−(−3.26329…)∣=0.65132…Δu9​=0.65132…
u10​=−2.04081…:Δu10​=0.57115…
f(u9​)=(−2.61197…)6+(−2.61197…)4−16(−2.61197…)3−(−2.61197…)2−1=641.38974…f′(u9​)=6(−2.61197…)5+4(−2.61197…)3−48(−2.61197…)2−2(−2.61197…)=−1122.97668…u10​=−2.04081…
Δu10​=∣−2.04081…−(−2.61197…)∣=0.57115…Δu10​=0.57115…
u11​=−1.54238…:Δu11​=0.49843…
f(u10​)=(−2.04081…)6+(−2.04081…)4−16(−2.04081…)3−(−2.04081…)2−1=220.42864…f′(u10​)=6(−2.04081…)5+4(−2.04081…)3−48(−2.04081…)2−2(−2.04081…)=−442.24518…u11​=−1.54238…
Δu11​=∣−1.54238…−(−2.04081…)∣=0.49843…Δu11​=0.49843…
u12​=−1.12448…:Δu12​=0.41790…
f(u11​)=(−1.54238…)6+(−1.54238…)4−16(−1.54238…)3−(−1.54238…)2−1=74.45277…f′(u11​)=6(−1.54238…)5+4(−1.54238…)3−48(−1.54238…)2−2(−1.54238…)=−178.15724…u12​=−1.12448…
Δu12​=∣−1.12448…−(−1.54238…)∣=0.41790…Δu12​=0.41790…
u13​=−0.80272…:Δu13​=0.32175…
f(u12​)=(−1.12448…)6+(−1.12448…)4−16(−1.12448…)3−(−1.12448…)2−1=24.10604…f′(u12​)=6(−1.12448…)5+4(−1.12448…)3−48(−1.12448…)2−2(−1.12448…)=−74.92024…u13​=−0.80272…
Δu13​=∣−0.80272…−(−1.12448…)∣=0.32175…Δu13​=0.32175…
u14​=−0.58368…:Δu14​=0.21904…
f(u13​)=(−0.80272…)6+(−0.80272…)4−16(−0.80272…)3−(−0.80272…)2−1=7.31448…f′(u13​)=6(−0.80272…)5+4(−0.80272…)3−48(−0.80272…)2−2(−0.80272…)=−33.39326…u14​=−0.58368…
Δu14​=∣−0.58368…−(−0.80272…)∣=0.21904…Δu14​=0.21904…
u15​=−0.46184…:Δu15​=0.12183…
f(u14​)=(−0.58368…)6+(−0.58368…)4−16(−0.58368…)3−(−0.58368…)2−1=1.99663…f′(u14​)=6(−0.58368…)5+4(−0.58368…)3−48(−0.58368…)2−2(−0.58368…)=−16.38770…u15​=−0.46184…
Δu15​=∣−0.46184…−(−0.58368…)∣=0.12183…Δu15​=0.12183…
u16​=−0.41933…:Δu16​=0.04251…
f(u15​)=(−0.46184…)6+(−0.46184…)4−16(−0.46184…)3−(−0.46184…)2−1=0.41814…f′(u15​)=6(−0.46184…)5+4(−0.46184…)3−48(−0.46184…)2−2(−0.46184…)=−9.83510…u16​=−0.41933…
Δu16​=∣−0.41933…−(−0.46184…)∣=0.04251…Δu16​=0.04251…
u17​=−0.41428…:Δu17​=0.00505…
f(u16​)=(−0.41933…)6+(−0.41933…)4−16(−0.41933…)3−(−0.41933…)2−1=0.04030…f′(u16​)=6(−0.41933…)5+4(−0.41933…)3−48(−0.41933…)2−2(−0.41933…)=−7.97447…u17​=−0.41428…
Δu17​=∣−0.41428…−(−0.41933…)∣=0.00505…Δu17​=0.00505…
u18​=−0.41421…:Δu18​=0.00006…
f(u17​)=(−0.41428…)6+(−0.41428…)4−16(−0.41428…)3−(−0.41428…)2−1=0.00052…f′(u17​)=6(−0.41428…)5+4(−0.41428…)3−48(−0.41428…)2−2(−0.41428…)=−7.76725…u18​=−0.41421…
Δu18​=∣−0.41421…−(−0.41428…)∣=0.00006…Δu18​=0.00006…
u19​=−0.41421…:Δu19​=1.19705E−8
f(u18​)=(−0.41421…)6+(−0.41421…)4−16(−0.41421…)3−(−0.41421…)2−1=9.29452E−8f′(u18​)=6(−0.41421…)5+4(−0.41421…)3−48(−0.41421…)2−2(−0.41421…)=−7.76450…u19​=−0.41421…
Δu19​=∣−0.41421…−(−0.41421…)∣=1.19705E−8Δu19​=1.19705E−8
u≈−0.41421…
Apply long division:u+0.41421…u6+u4−16u3−u2−1​=u5−0.41421…u4+1.17157…u3−16.48528…u2+5.82842…u−2.41421…
u5−0.41421…u4+1.17157…u3−16.48528…u2+5.82842…u−2.41421…≈0
Find one solution for u5−0.41421…u4+1.17157…u3−16.48528…u2+5.82842…u−2.41421…=0 using Newton-Raphson:u≈2.41421…
u5−0.41421…u4+1.17157…u3−16.48528…u2+5.82842…u−2.41421…=0
Newton-Raphson Approximation Definition
f(u)=u5−0.41421…u4+1.17157…u3−16.48528…u2+5.82842…u−2.41421…
Find f′(u):5u4−1.65685…u3+3.51471…u2−32.97056…u+5.82842…
dud​(u5−0.41421…u4+1.17157…u3−16.48528…u2+5.82842…u−2.41421…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(u5)−dud​(0.41421…u4)+dud​(1.17157…u3)−dud​(16.48528…u2)+dud​(5.82842…u)−dud​(2.41421…)
dud​(u5)=5u4
dud​(u5)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=5u5−1
Simplify=5u4
dud​(0.41421…u4)=1.65685…u3
dud​(0.41421…u4)
Take the constant out: (a⋅f)′=a⋅f′=0.41421…dud​(u4)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=0.41421…⋅4u4−1
Simplify=1.65685…u3
dud​(1.17157…u3)=3.51471…u2
dud​(1.17157…u3)
Take the constant out: (a⋅f)′=a⋅f′=1.17157…dud​(u3)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=1.17157…⋅3u3−1
Simplify=3.51471…u2
dud​(16.48528…u2)=32.97056…u
dud​(16.48528…u2)
Take the constant out: (a⋅f)′=a⋅f′=16.48528…dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=16.48528…⋅2u2−1
Simplify=32.97056…u
dud​(5.82842…u)=5.82842…
dud​(5.82842…u)
Take the constant out: (a⋅f)′=a⋅f′=5.82842…dudu​
Apply the common derivative: dudu​=1=5.82842…⋅1
Simplify=5.82842…
dud​(2.41421…)=0
dud​(2.41421…)
Derivative of a constant: dxd​(a)=0=0
=5u4−1.65685…u3+3.51471…u2−32.97056…u+5.82842…−0
Simplify=5u4−1.65685…u3+3.51471…u2−32.97056…u+5.82842…
Let u0​=0Compute un+1​ until Δun+1​<0.000001
u1​=0.41421…:Δu1​=0.41421…
f(u0​)=05−0.41421…⋅04+1.17157…⋅03−16.48528…⋅02+5.82842…⋅0−2.41421…=−2.41421…f′(u0​)=5⋅04−1.65685…⋅03+3.51471…⋅02−32.97056…⋅0+5.82842…=5.82842…u1​=0.41421…
Δu1​=∣0.41421…−0∣=0.41421…Δu1​=0.41421…
u2​=0.03272…:Δu2​=0.38148…
f(u1​)=0.41421…5−0.41421…⋅0.41421…4+1.17157…⋅0.41421…3−16.48528…⋅0.41421…2+5.82842…⋅0.41421…−2.41421…=−2.74516…f′(u1​)=5⋅0.41421…4−1.65685…⋅0.41421…3+3.51471…⋅0.41421…2−32.97056…⋅0.41421…+5.82842…=−7.19595…u2​=0.03272…
Δu2​=∣0.03272…−0.41421…∣=0.38148…Δu2​=0.38148…
u3​=0.50422…:Δu3​=0.47149…
f(u2​)=0.03272…5−0.41421…⋅0.03272…4+1.17157…⋅0.03272…3−16.48528…⋅0.03272…2+5.82842…⋅0.03272…−2.41421…=−2.24108…f′(u2​)=5⋅0.03272…4−1.65685…⋅0.03272…3+3.51471…⋅0.03272…2−32.97056…⋅0.03272…+5.82842…=4.75313…u3​=0.50422…
Δu3​=∣0.50422…−0.03272…∣=0.47149…Δu3​=0.47149…
u4​=0.14569…:Δu4​=0.35852…
f(u3​)=0.50422…5−0.41421…⋅0.50422…4+1.17157…⋅0.50422…3−16.48528…⋅0.50422…2+5.82842…⋅0.50422…−2.41421…=−3.51061…f′(u3​)=5⋅0.50422…4−1.65685…⋅0.50422…3+3.51471…⋅0.50422…2−32.97056…⋅0.50422…+5.82842…=−9.79171…u4​=0.14569…
Δu4​=∣0.14569…−0.50422…∣=0.35852…Δu4​=0.35852…
u5​=1.88887…:Δu5​=1.74318…
f(u4​)=0.14569…5−0.41421…⋅0.14569…4+1.17157…⋅0.14569…3−16.48528…⋅0.14569…2+5.82842…⋅0.14569…−2.41421…=−1.91147…f′(u4​)=5⋅0.14569…4−1.65685…⋅0.14569…3+3.51471…⋅0.14569…2−32.97056…⋅0.14569…+5.82842…=1.09654…u5​=1.88887…
Δu5​=∣1.88887…−0.14569…∣=1.74318…Δu5​=1.74318…
u6​=4.63637…:Δu6​=2.74749…
f(u5​)=1.88887…5−0.41421…⋅1.88887…4+1.17157…⋅1.88887…3−16.48528…⋅1.88887…2+5.82842…⋅1.88887…−2.41421…=−23.55475…f′(u5​)=5⋅1.88887…4−1.65685…⋅1.88887…3+3.51471…⋅1.88887…2−32.97056…⋅1.88887…+5.82842…=8.57317…u6​=4.63637…
Δu6​=∣4.63637…−1.88887…∣=2.74749…Δu6​=2.74749…
u7​=3.79830…:Δu7​=0.83806…
f(u6​)=4.63637…5−0.41421…⋅4.63637…4+1.17157…⋅4.63637…3−16.48528…⋅4.63637…2+5.82842…⋅4.63637…−2.41421…=1737.96023…f′(u6​)=5⋅4.63637…4−1.65685…⋅4.63637…3+3.51471…⋅4.63637…2−32.97056…⋅4.63637…+5.82842…=2073.76688…u7​=3.79830…
Δu7​=∣3.79830…−4.63637…∣=0.83806…Δu7​=0.83806…
u8​=3.17364…:Δu8​=0.62465…
f(u7​)=3.79830…5−0.41421…⋅3.79830…4+1.17157…⋅3.79830…3−16.48528…⋅3.79830…2+5.82842…⋅3.79830…−2.41421…=550.45789…f′(u7​)=5⋅3.79830…4−1.65685…⋅3.79830…3+3.51471…⋅3.79830…2−32.97056…⋅3.79830…+5.82842…=881.21646…u8​=3.17364…
Δu8​=∣3.17364…−3.79830…∣=0.62465…Δu8​=0.62465…
u9​=2.74529…:Δu9​=0.42835…
f(u8​)=3.17364…5−0.41421…⋅3.17364…4+1.17157…⋅3.17364…3−16.48528…⋅3.17364…2+5.82842…⋅3.17364…−2.41421…=167.42483…f′(u8​)=5⋅3.17364…4−1.65685…⋅3.17364…3+3.51471…⋅3.17364…2−32.97056…⋅3.17364…+5.82842…=390.85896…u9​=2.74529…
Δu9​=∣2.74529…−3.17364…∣=0.42835…Δu9​=0.42835…
u10​=2.50516…:Δu10​=0.24012…
f(u9​)=2.74529…5−0.41421…⋅2.74529…4+1.17157…⋅2.74529…3−16.48528…⋅2.74529…2+5.82842…⋅2.74529…−2.41421…=45.99071…f′(u9​)=5⋅2.74529…4−1.65685…⋅2.74529…3+3.51471…⋅2.74529…2−32.97056…⋅2.74529…+5.82842…=191.52771…u10​=2.50516…
Δu10​=∣2.50516…−2.74529…∣=0.24012…Δu10​=0.24012…
u11​=2.42337…:Δu11​=0.08179…
f(u10​)=2.50516…5−0.41421…⋅2.50516…4+1.17157…⋅2.50516…3−16.48528…⋅2.50516…2+5.82842…⋅2.50516…−2.41421…=9.50260…f′(u10​)=5⋅2.50516…4−1.65685…⋅2.50516…3+3.51471…⋅2.50516…2−32.97056…⋅2.50516…+5.82842…=116.17302…u11​=2.42337…
Δu11​=∣2.42337…−2.50516…∣=0.08179…Δu11​=0.08179…
u12​=2.41431…:Δu12​=0.00905…
f(u11​)=2.42337…5−0.41421…⋅2.42337…4+1.17157…⋅2.42337…3−16.48528…⋅2.42337…2+5.82842…⋅2.42337…−2.41421…=0.86400…f′(u11​)=5⋅2.42337…4−1.65685…⋅2.42337…3+3.51471…⋅2.42337…2−32.97056…⋅2.42337…+5.82842…=95.43428…u12​=2.41431…
Δu12​=∣2.41431…−2.42337…∣=0.00905…Δu12​=0.00905…
u13​=2.41421…:Δu13​=0.00010…
f(u12​)=2.41431…5−0.41421…⋅2.41431…4+1.17157…⋅2.41431…3−16.48528…⋅2.41431…2+5.82842…⋅2.41431…−2.41421…=0.00977…f′(u12​)=5⋅2.41431…4−1.65685…⋅2.41431…3+3.51471…⋅2.41431…2−32.97056…⋅2.41431…+5.82842…=93.27961…u13​=2.41421…
Δu13​=∣2.41421…−2.41431…∣=0.00010…Δu13​=0.00010…
u14​=2.41421…:Δu14​=1.39211E−8
f(u13​)=2.41421…5−0.41421…⋅2.41421…4+1.17157…⋅2.41421…3−16.48528…⋅2.41421…2+5.82842…⋅2.41421…−2.41421…=1.29821E−6f′(u13​)=5⋅2.41421…4−1.65685…⋅2.41421…3+3.51471…⋅2.41421…2−32.97056…⋅2.41421…+5.82842…=93.25483…u14​=2.41421…
Δu14​=∣2.41421…−2.41421…∣=1.39211E−8Δu14​=1.39211E−8
u≈2.41421…
Apply long division:u−2.41421…u5−0.41421…u4+1.17157…u3−16.48528…u2+5.82842…u−2.41421…​=u4+2u3+6u2−2u+1
u4+2u3+6u2−2u+1≈0
Find one solution for u4+2u3+6u2−2u+1=0 using Newton-Raphson:No Solution for u∈R
u4+2u3+6u2−2u+1=0
Newton-Raphson Approximation Definition
f(u)=u4+2u3+6u2−2u+1
Find f′(u):4u3+6u2+12u−2
dud​(u4+2u3+6u2−2u+1)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(u4)+dud​(2u3)+dud​(6u2)−dud​(2u)+dud​(1)
dud​(u4)=4u3
dud​(u4)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=4u4−1
Simplify=4u3
dud​(2u3)=6u2
dud​(2u3)
Take the constant out: (a⋅f)′=a⋅f′=2dud​(u3)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=2⋅3u3−1
Simplify=6u2
dud​(6u2)=12u
dud​(6u2)
Take the constant out: (a⋅f)′=a⋅f′=6dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=6⋅2u2−1
Simplify=12u
dud​(2u)=2
dud​(2u)
Take the constant out: (a⋅f)′=a⋅f′=2dudu​
Apply the common derivative: dudu​=1=2⋅1
Simplify=2
dud​(1)=0
dud​(1)
Derivative of a constant: dxd​(a)=0=0
=4u3+6u2+12u−2+0
Simplify=4u3+6u2+12u−2
Let u0​=1Compute un+1​ until Δun+1​<0.000001
u1​=0.6:Δu1​=0.4
f(u0​)=14+2⋅13+6⋅12−2⋅1+1=8f′(u0​)=4⋅13+6⋅12+12⋅1−2=20u1​=0.6
Δu1​=∣0.6−1∣=0.4Δu1​=0.4
u2​=0.29338…:Δu2​=0.30661…
f(u1​)=0.64+2⋅0.63+6⋅0.62−2⋅0.6+1=2.5216f′(u1​)=4⋅0.63+6⋅0.62+12⋅0.6−2=8.224u2​=0.29338…
Δu2​=∣0.29338…−0.6∣=0.30661…Δu2​=0.30661…
u3​=−0.16852…:Δu3​=0.46190…
f(u2​)=0.29338…4+2⋅0.29338…3+6⋅0.29338…2−2⋅0.29338…+1=0.98759…f′(u2​)=4⋅0.29338…3+6⋅0.29338…2+12⋅0.29338…−2=2.13808…u3​=−0.16852…
Δu3​=∣−0.16852…−0.29338…∣=0.46190…Δu3​=0.46190…
u4​=0.21863…:Δu4​=0.38715…
f(u3​)=(−0.16852…)4+2(−0.16852…)3+6(−0.16852…)2−2(−0.16852…)+1=1.49867…f′(u3​)=4(−0.16852…)3+6(−0.16852…)2+12(−0.16852…)−2=−3.87099…u4​=0.21863…
Δu4​=∣0.21863…−(−0.16852…)∣=0.38715…Δu4​=0.38715…
u5​=−0.69789…:Δu5​=0.91652…
f(u4​)=0.21863…4+2⋅0.21863…3+6⋅0.21863…2−2⋅0.21863…+1=0.87272…f′(u4​)=4⋅0.21863…3+6⋅0.21863…2+12⋅0.21863…−2=0.95220…u5​=−0.69789…
Δu5​=∣−0.69789…−0.21863…∣=0.91652…Δu5​=0.91652…
u6​=−0.14461…:Δu6​=0.55328…
f(u5​)=(−0.69789…)4+2(−0.69789…)3+6(−0.69789…)2−2(−0.69789…)+1=4.87554…f′(u5​)=4(−0.69789…)3+6(−0.69789…)2+12(−0.69789…)−2=−8.81206…u6​=−0.14461…
Δu6​=∣−0.14461…−(−0.69789…)∣=0.55328…Δu6​=0.55328…
u7​=0.24442…:Δu7​=0.38903…
f(u6​)=(−0.14461…)4+2(−0.14461…)3+6(−0.14461…)2−2(−0.14461…)+1=1.40910…f′(u6​)=4(−0.14461…)3+6(−0.14461…)2+12(−0.14461…)−2=−3.62201…u7​=0.24442…
Δu7​=∣0.24442…−(−0.14461…)∣=0.38903…Δu7​=0.38903…
u8​=−0.42403…:Δu8​=0.66846…
f(u7​)=0.24442…4+2⋅0.24442…3+6⋅0.24442…2−2⋅0.24442…+1=0.90238…f′(u7​)=4⋅0.24442…3+6⋅0.24442…2+12⋅0.24442…−2=1.34994…u8​=−0.42403…
Δu8​=∣−0.42403…−0.24442…∣=0.66846…Δu8​=0.66846…
u9​=0.02045…:Δu9​=0.44448…
f(u8​)=(−0.42403…)4+2(−0.42403…)3+6(−0.42403…)2−2(−0.42403…)+1=2.80677…f′(u8​)=4(−0.42403…)3+6(−0.42403…)2+12(−0.42403…)−2=−6.31459…u9​=0.02045…
Δu9​=∣0.02045…−(−0.42403…)∣=0.44448…Δu9​=0.44448…
u10​=0.56930…:Δu10​=0.54885…
f(u9​)=0.02045…4+2⋅0.02045…3+6⋅0.02045…2−2⋅0.02045…+1=0.96162…f′(u9​)=4⋅0.02045…3+6⋅0.02045…2+12⋅0.02045…−2=−1.75205…u10​=0.56930…
Δu10​=∣0.56930…−0.02045…∣=0.54885…Δu10​=0.54885…
Cannot find solution
The solutions areu≈−0.41421…,u≈2.41421…
The solutions areu=−1,u=1,u≈−0.41421…,u≈2.41421…
u=−1,u=1,u≈−0.41421…,u≈2.41421…
Verify Solutions
Find undefined (singularity) points:u=0
Take the denominator(s) of u4+u−4 and compare to zero
Solve u4=0:u=0
u4=0
Apply rule xn=0⇒x=0
u=0
Take the denominator(s) of 16(u−u−1)+2 and compare to zero
u=0
The following points are undefinedu=0
Combine undefined points with solutions:
u=−1,u=1,u≈−0.41421…,u≈2.41421…
u=−1,u=1,u≈−0.41421…,u≈2.41421…
Substitute back u=ex,solve for x
Solve ex=−1:No Solution for x∈R
ex=−1
af(x) cannot be zero or negative for x∈RNoSolutionforx∈R
Solve ex=1:x=0
ex=1
Apply exponent rules
ex=1
If f(x)=g(x), then ln(f(x))=ln(g(x))ln(ex)=ln(1)
Apply log rule: ln(ea)=aln(ex)=xx=ln(1)
Simplify ln(1):0
ln(1)
Apply log rule: loga​(1)=0=0
x=0
x=0
Solve ex=−0.41421…:No Solution for x∈R
ex=−0.41421…
af(x) cannot be zero or negative for x∈RNoSolutionforx∈R
Solve ex=2.41421…:x=ln(2.41421…)
ex=2.41421…
Apply exponent rules
ex=2.41421…
If f(x)=g(x), then ln(f(x))=ln(g(x))ln(ex)=ln(2.41421…)
Apply log rule: ln(ea)=aln(ex)=xx=ln(2.41421…)
x=ln(2.41421…)
x=0,x=ln(2.41421…)
x=0,x=ln(2.41421…)

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Popular Examples

cos(θ)=((sqrt(3))/2)cos(x+(3pi)/4)+cos(x-(3pi)/4)=15-sin(θ)=cos(2θ)8sin(θ)+4sqrt(3)=0solvefor x,sin(x)=(sqrt(3))/2

Frequently Asked Questions (FAQ)

  • What is the general solution for cosh(4x)=16sinh(x)+1 ?

    The general solution for cosh(4x)=16sinh(x)+1 is x=0,x=ln(2.41421…)
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