How do you use the chain rule to differentiate #f(x)=sqrt(sinx+x^-9+6x^-20)#?

1 Answer
Tazwar Sikder
May 18, 2017

# f'(x) = frac(1)(2) sqrt(sin(x) + x^(- 9) + 6 x^(- 20)) (cos(x) - frac(9)(x^(10)) - frac(120)(x^(21)))#

Explanation:

We have: #f(x) = sqrt(sin(x) + x^(- 9) + 6 x^(- 20))#

Let #u = sin(x) + x^(- 9) + 6 x^(- 20)# and #v = sqrt(u) = u^(frac(1)(2))#:

#Rightarrow f'(x) = u' cdot v'#

#Rightarrow f'(x) = frac(1)(2)u^(frac(1)(2) - 1) cdot (cos(x) - 9 x^(- 9 - 1) + 6 cdot (- 20) x^(- 20 - 1))#

#Rightarrow f'(x) = frac(1)(2)u^(- frac(1)(2)) cdot (cos(x) - 9 x^(- 10) - 120 x^(- 21))#

#Rightarrow f'(x) = frac(1)(2) cos(x) sqrt(u) - frac(9)(2) x^(- 10) sqrt(u) - 60 x^(- 21) sqrt(u)#

Now, let's replace #u# with #sin(x) + x^(- 9) + 6 x^(- 20)#:

#Rightarrow f'(x) = frac(1)(2) cos(x) sqrt(sin(x) + x^(- 9) + 6 x^(- 20)) - frac(9)(2) x^(- 10) sqrt(sin(x) + x^(- 9) + 6 x^(- 20)) - 60 x^(- 21) sqrt(sin(x) + x^(- 9) + 6 x^(- 20))#

#Rightarrow f'(x) = frac(1)(2) sqrt(sin(x) + x^(- 9) + 6 x^(- 20)) (cos(x) - frac(9)(x^(10)) - frac(120)(x^(21)))#

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