Question

How do you differentiate `f(x)=tan(sqrt(x^3-1))` using the chain rule?

Answer 1

`f'(x)= (3x^2sec^2(sqrt(x^3-1)))/(2sqrt(x^3-1)`

Explanation:

`f(x)=tan(sqrt(x^3-1))`

To find `f'`, set `u=sqrt(x^3-1)` and `u' = (3x^2)/(2sqrt(x^3-1))` and `f'(u) = sec^2u=sec^2(sqrt(x^3-1))`so that `f'(x) = f'(u) xx u'=(3x^2sec^2(sqrt(x^3-1)))/(2sqrt(x^3-1)`

Answer 2

`f'(x)=(3x^2sec^2(sqrt(x^3-1)))/(2sqrt(x^3-1))`

Explanation:

`"differentiate using the "color(blue)"chain rule"`

`"given " y=f(g(h(x)))" then"`

`dy/dx=f'(g(h(x))).g'(h(x)).h'(x)larr" chain rule"`

`y=tan(sqrt(x^3-1))`

`rArrf'(g(h(x)))=sec^2(sqrt(x^3-1))to(color(red)(1))`

`g(h(x))=(x^3-1)^(1/2)rArrg'(h(x))=1/2(x^3-1)^(-1/2)to(color(red)(2))`

`h(x)=x^3-1rArrh'(x)=3x^2to(color(red)(3))`

`"combining the product of all 3 parts gives"`

`rArrf'(x)=dy/dx=(3x^2sec^2(sqrt(x^3-1)))/(2sqrt(x^3-1))`