derivative of e^{ln(cos(x^2)}sqrt(2x+1))

Lösung

\frac{d}{d x} (e^{l n (c o s (x^{2}))} 2 x + 1)

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

Schritte zur Lösung

\frac{d}{d x} (e^{l n (c o s (x^{2}))} 2 x + 1)

Vereinfache

e^{l n (c o s (x^{2}))} 2 x + 1 : cos (x^{2}) 2 x + 1

= \frac{d}{d x} (cos (x^{2}) 2 x + 1)

Wende die Produktregel an:

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

f = cos (x^{2}), g = 2 x + 1

= \frac{d}{d x} (cos (x^{2})) 2 x + 1 + \frac{d}{d x} (2 x + 1) cos (x^{2})

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

\frac{d}{d x} (2 x + 1) = \frac{1}{2 x + 1}

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

Vereinfache

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

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

\frac{d y}{d x} + 2 x y = 5 x

s l o p e in t erce pt (- 4, - 6), (- 2, 0)

n = 1 \sum \infty n^{5} (1 + n^{6})^{6}

\frac{d}{d x} (\frac{f ( x )}{x ^{5}})

d er i v a t i v e (5 x + 2)^{4} (2 x - 3)^{8}