derivative of tan(cot(ln(x))+sec(e^x))

Lösung

\frac{d}{d x} (tan (cot (ln (x))) + sec (e^{x}))

- \frac{csc ^{2} ( ln ( x ) ) sec ^{2} ( cot ( ln ( x ) ) )}{x} + sec (e^{x}) tan (e^{x}) e^{x}

Schritte zur Lösung

\frac{d}{d x} (tan (cot (ln (x))) + sec (e^{x}))

Wende die Summen-/Differenzregel an:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (tan (cot (ln (x)))) + \frac{d}{d x} (sec (e^{x}))

\frac{d}{d x} (tan (cot (ln (x)))) = - \frac{csc ^{2} ( ln ( x ) ) sec ^{2} ( cot ( ln ( x ) ) )}{x}

\frac{d}{d x} (sec (e^{x})) = sec (e^{x}) tan (e^{x}) e^{x}

= - \frac{csc ^{2} ( ln ( x ) ) sec ^{2} ( cot ( ln ( x ) ) )}{x} + sec (e^{x}) tan (e^{x}) e^{x}

x \to 0 + lim (\frac{sin ( x )}{x})

\int 4 x^{5} e^{x^{6}} d x

\frac{\partial}{\partial y} (6 x + 10 y)

\frac{d}{d x} (6 (4 e^{x^{2}} x^{3} + 6 e^{x^{2}} x))

5 x^{^{''}} + 5 x^{^{'}} + 6 x = 0, x^{^{'}} (0) = 0.45, x (0) = - 0.3