What is the derivative of \sqrt[4]{cos(e^{6x}}) ?

The derivative of \sqrt[4]{cos(e^{6x}}) is -(3e^{6x}sin(e^{6x}))/(2cos^{3/4)(e^{6x})}

What is the first derivative of \sqrt[4]{cos(e^{6x}}) ?

The first derivative of \sqrt[4]{cos(e^{6x}}) is -(3e^{6x}sin(e^{6x}))/(2cos^{3/4)(e^{6x})}

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- \frac{3 e ^{6 x} sin ( e ^{6 x} )}{2 cos ^{\frac{3}{4}} ( e ^{6 x} )}

Solution steps

Apply the chain rule:

\frac{1}{4 ( cos ( e ^{6 x} ) ) ^{\frac{3}{4}}} \frac{d}{d x} (cos (e^{6 x}))

= \frac{1}{4 ( cos ( e ^{6 x} ) ) ^{\frac{3}{4}}} \frac{d}{d x} (cos (e^{6 x}))

\frac{d}{d x} (cos (e^{6 x})) = - sin (e^{6 x}) e^{6 x} \cdot 6

= \frac{1}{4 ( cos ( e ^{6 x} ) ) ^{\frac{3}{4}}} (- sin (e^{6 x}) e^{6 x} \cdot 6)

Simplify

\frac{1}{4 cos ^{\frac{3}{4}} ( e ^{6 x} )} (- sin (e^{6 x}) e^{6 x} \cdot 6) : - \frac{3 e ^{6 x} sin ( e ^{6 x} )}{2 cos ^{\frac{3}{4}} ( e ^{6 x} )}

= - \frac{3 e ^{6 x} sin ( e ^{6 x} )}{2 cos ^{\frac{3}{4}} ( e ^{6 x} )}

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

sin (x), \frac{1}{2}

f (x) = tan^{40} (sec (cos (x)))

\int 3 sin^{3} (x) d x

y^{^{''}} + 16 y = - 32 sin (4 t)

What is the derivative of \sqrt[4]{cos(e^{6x}}) ?
The derivative of \sqrt[4]{cos(e^{6x}}) is -(3e^{6x}sin(e^{6x}))/(2cos^{3/4)(e^{6x})}
What is the first derivative of \sqrt[4]{cos(e^{6x}}) ?
The first derivative of \sqrt[4]{cos(e^{6x}}) is -(3e^{6x}sin(e^{6x}))/(2cos^{3/4)(e^{6x})}