# DerivativeFind the first derivative of y=tan^4(x+1)^4

### 2 Answers | Add Yours

You need use implicit differentiation, such that:

`(dy)/(dx) = (d(tan^4(x+1)^4))/(dx)`

`(dy)/(dx) = 4tan^3(x+1)^4*(d(tan(x+1)^4))/(dx)`

`(dy)/(dx) = 4tan^3(x+1)^4*(d(tan(x+1)^4))/(dx)`

`(dy)/(dx) = 4tan^3(x+1)^4*(1/(cos^2(x + 1)^4))*(d((x+1)^4))/(dx)`

`(dy)/(dx) = 4tan^3(x+1)^4*(1/(cos^2(x + 1)^4))*4(x+1)^3*(d(x+1))/(dx)`

`(dy)/(dx) = 4tan^3(x+1)^4*(1/(cos^2(x + 1)^4))*4(x+1)^3*1`

`(dy)/(dx) = 16(((x+1)tan(x+1)^4)^3)/(cos^2(x + 1)^4)`

**Hence, evaluating the derivative of the given function, yields `(dy)/(dx) = 16(((x+1)tan(x+1)^4)^3)/(cos^2(x + 1)^4).` **

f(x) = [tan(x+1)^4]^4

We'll use the chain rule to differentiate the function.

First, we'll differentiate the power function. We'll put

tan v(x) = u(v(x))

(u^4)' = 4u^3

{[tan v(x)]^4}' = 4 [tan v(x)]^3*v'(x)

We'll put v(x) = (x+1)^4

v'(x) = 4(x+1)^3*(x+1)'

v'(x) = 4(x+1)^3

f'(x) = 4 [tan v(x)]^3*v'(x)

We'll replace v(x) and v'(x) and we'll get:

f'(x) = 4 [tan(x+1)^4]^3*[4(x+1)^3]

**f'(x) = 16[tan(x+1)^4]^3*[(x+1)^3]**