derivative f(x)=cos(x^2)*2x

Lösung

ableitung von

f (x) = cos (x^{2}) \cdot 2 x

2 (- 2 x^{2} sin (x^{2}) + cos (x^{2}))

Schritte zur Lösung

\frac{d}{d x} (cos (x^{2}) \cdot 2 x)

Entferne die Konstante:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d}{d x} (cos (x^{2}) x)

Wende die Produktregel an:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = cos (x^{2}), g = x

= 2 (\frac{d}{d x} (cos (x^{2})) x + \frac{d x}{d x} cos (x^{2}))

\frac{d}{d x} (cos (x^{2})) = - sin (x^{2}) \cdot 2 x

\frac{d x}{d x} = 1

= 2 ((- sin (x^{2}) \cdot 2 x) x + 1 \cdot cos (x^{2}))

Vereinfache

2 ((- sin (x^{2}) \cdot 2 x) x + 1 \cdot cos (x^{2})) : 2 (- 2 x^{2} sin (x^{2}) + cos (x^{2}))

= 2 (- 2 x^{2} sin (x^{2}) + cos (x^{2}))

\int \frac{15}{16 + r ^{2}} d r

l a pl a ce t r an s f or m 6 e^{- 4 t} + 5 e^{- 5 t} + 4 t e^{- 6 t}

\frac{d}{d x} (4 (x^{2} + 5 x + 2) tan (- x))

\frac{\partial}{\partial x} (2 x^{0.5} + y)

(x + 1) \frac{d y}{d x} + x y = ln (x)