What is the derivative of (cosx)^sin2x?

Expert Answers

An illustration of the letter 'A' in a speech bubbles

In this question, the sin (2x) is a power of the base, cos x.

`cos x ^(sin 2x)`

To find the derivative, the natural logarithm function is used.

Let y equal cosx ^ (sin 2x).


Take the natural logarithm of both sides.

`ln [y] = ln[cosx^((sin2x))]`

Use the logarithm property that ln[x^b] = b*ln[x].


Take the derivative of both sides using implicit differentiation and the product rule.

`1/y * dy/dx=(sin2x)*1/cosx*(-sinx)+ln[cosx]*cos(2x)*2`

Multiply both sides by y.


Replace y with the cosx^(sin2x).


The derivative can be simplified in many different forms. My suggestion is have sin/cos become tangent.

One simplified form of the derivative is:



Approved by eNotes Editorial Team
An illustration of the letter 'A' in a speech bubbles

we have to find d(cosx * sin2x)/dx.

This can be simply done by using the principle of deriavation of multiplications.

d(cosx * sin2x)/dx = cosx * d(sin2x)/dx + sin2x * d(cosx)/dx

= cosx*2*cos2x+ sin2x*(-sinx)

= 2*cosx*cos2x - sinx*sin2x

=2*cosx*cos2x - sinx*2*sinx*cosx (sin2x = 2*sinx*cosx)

= 2*cosx* [cos2x-(sinx)^2]

= 2*cosx*[(cosx)^2 - (sinx)^2-(sinx)^2]

(cos2x = (cosx)^2 - (sinx)^2)

d(cosx * sin2x)/dx= 2*cosx[(cosx)^2 - 2*(sinx)^2]

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial Team