Question

How do you differentiate `y = x^(2 cos x)`?

Answer 1

The derivative for this is found by the chain rule.

Explanation:

Let, `y=f(x)=x^(2cosx)`.

`:.dy/dx=d/(dx)(x^(2cosx))*d/(dx)(2cosx)`.

`:.dy/dx=(2cosx*x^(2cosx-1))*(-2sinx)`.

`:.dy/dx=-4sinxcosx*x^(2cosx-1)`. (answer).

Answer 2

Use some version of logarithmic differentiation.

Explanation:

My current preferred form for logarithmic differfentiation is to rewrite as `e` to a power.

`y = x^(2cosx) = e^(2cosxlnx)`

Now use `d/dx(e^u) = e^u (du)/dx` to get

`dy/dx = e^(2cosxlnx) * d/dx(2cosxlnx)`

`= x^(2cosx) *[(2cosx)/x - 2sinxlnx]`

Answer 3

Please see the explanation for the steps leading to:

`dy/dx = 2(cos(x)/x - sin(x)ln(x))x^(2cos(x))`

Explanation:

Let `ln(y) = ln(x^(2cos(x)))`

`ln(y) = 2cos(x)ln(x)`

`1/y(dy/dx) = -2sin(x)ln(x) + 2cos(x)/x`

`dy/dx = 2(cos(x)/x - sin(x)ln(x))y`

Substitute `x^(2cos(x))` for y:

`dy/dx = 2(cos(x)/x - sin(x)ln(x))x^(2cos(x))`