Question

Differentiate and simply and find the equation. Please read the image and help ?!

Answer 1

`(df)/(dx)=(8x^3)/(x^8-2x^4+2)` and tangent at `(1,0)` is `y=8x-8`

Explanation:

Derivative of `tan^(-1)x` is `1/(1+x^2)`

Now we have `f(x)=2tan^(-1)(x^4-1)`

Hence `f'(x)=(df)/(dx)=2/(1+(x^4-1)^2)*4x^3`

= `(8x^3)/(1+(x^4-1)^2)`

= `(8x^3)/(1+x^8+1-2x^4)=(8x^3)/(x^8-2x^4+2)`

and as slope of tangent at `(x_0,f(x_0))` is given by `f'(x_0)`

hence slope of tangent at `(1,0)` is

`f'(1)=(8*1)/(1+(1^4-1)^2)=8`

and hence using point slope form of equation,

equation of tangent is

`(y-0)=8(x-1)` i.e. `y=8x-8`