Question

How do I differentiate `y=(tanx)^(x^2)` using natural logarithms?

Answer 1

`dy/dx = (x^2secxcscx+2xlntanx)(tanx)^(x^2)`

Explanation:

`y = (tanx)^(x^2)`

`lny = x^2lntanx`

`1/ydy/dx = x^2 sec^2x/tanx +2xlntanx=x^2secxcscx+2xlntanx`

`dy/dx = y(x^2secxcscx+2xlntanx) = (x^2secxcscx+2xlntanx)(tanx)^(x^2)`

Answer 2

`dy/dx=(2x(ln(tanx))+x^2(sec^2x/tanx))*(tanx)^(x^2)`

Explanation:

to diffrentiate, we usually take:
`y=(tanx)^(x^2)`
`d/dxy=d/dx(tanx)^(x^2)`

But this time since x is in the power, we are going to use log rule before starting with differentiation:

We begin by taking `ln`(natural log) of both sides:

`lny=ln(tan^x)^(x^2)`

applying the log rule:
`loga^x=x*loga`
we get:
`lny=x^2ln(tanx)`

Differentiating both sides with respect to `x`
`d/(dx)lny=d/(dx)x^2ln(tanx)`

Using implicit differentiation's rule (`d/dx f(y)=d/dy (f(y) *dy/dx`):

`d/dylny*d/(dx)y=1/y*dy/dx` (since derrivative of 1/x=x)

`rarr 1/y*d/dxy= d/(dx)x^2ln(tanx)`

multiplying both sides by y to isolate dy/dx:
`1/y*d/dxy*y=(d/(dx)x^2ln(tanx))*y`
`rarrdy/dx=(d/(dx)x^2ln(tanx))*y`

we use chain rule to differentiate the right hand side:
`dy/dx=(d/(dx)(x^2)(ln(tanx))+x^2(d/dx(ln(tanx)))*y`
`dy/dx=(2x(ln(tanx))+x^2(1/tanx*sec^2x))*y`
`dy/dx=(2x(ln(tanx))+x^2(sec^2x/tanx))*y`

substituting y by its value in terms of x from given:
`y=(tanx)^(x^2)`
`rarr dy/dx=(2x(ln(tanx))+x^2(sec^2x/tanx))*(tanx)^(x^2)`

simplifying the answer will give you:
`2x*(tanx)^(x^2)*ln(tanx)+x(csc2x)`