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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