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

2 Answers
Monzur R.
Jul 20, 2017

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)

Jim G.
Jul 20, 2017

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))

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