What is the derivative of (cosx)^sin2x?

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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]**

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).

`y=cosx^((sin2x))`

Take the natural logarithm of both sides.

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

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

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

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.

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

Replace y with the cosx^(sin2x).

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

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

One simplified form of the derivative is:

`d/dx(cosx^(sin2x))=cosx^(sin2x)[-(sin2x)(tanx)+2(cos2x)ln[cosx]]`

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