`f(s) = sqrt((s^2 + 1)/(s^2 + 4))` Find the derivative of the function.

Textbook Question

Chapter 3, 3.4 - Problem 22 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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`f(s)=sqrt((s^2+1)/(s^2+4))`

We can find derivative either by product rule or by quotient rule.

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

Using product rule

`f'(s)=(s^2+1)^(1/2) * d/(ds) (s^2+4)^(-1/2) + (s^2+4)^(-1/2) * d/(ds) (s^2+1)^(1/2)`

`f'(s)=(s^2+1)^(1/2)*(-1/2)(s^2+4)^(-3/2) (2s)+ (s^2+4)^(-1/2)*(1/2)(s^2+1)^(-1/2)(2s)`

`f'(s)= (-s(s^2+1)^(1/2))/((s^2+4)^(3/2)) + (s)/(sqrt(s^2+4)sqrt(s^2+1))`

`f'(s)=(-ssqrt(s^2+1))/((s^2+4)sqrt(s^2+4)) + s/(sqrt(s^2+4)sqrt(s^2+1))`

`f'(s)=(-s(s^2+1)sqrt(s^2+4) + s(s^2+4)sqrt(s^2+4))/((s^2+4)^2sqrt(s^2+1))`

`f'(s)= (ssqrt(s^2+4)*(-s^(2)-1+s^(2)+4))/((s^2+4)^2sqrt(s^2+1))`

`f'(s)=(3ssqrt(s^2+4))/((s^2+4)^2sqrt(s^2+1))`

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hkj1385 | (Level 1) Assistant Educator

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Note:- 1) If y = x^n ; then dy/dx = n*{x^(n-1)}

2) If y = sqrt(x) ; then dy/dx = 1/{2*sqrt(x)}

3) If a function to be differentiated contains sub-functions,then by the rule of differentiation, the last function is differentiated first.

4) If the function is of the form y = u/v ; where u & v are both functions of 'x' , then dy/dx = y' = [{v*u' - u*v'}/(v^2)]

Now, for the given question , find the solution in the attachment

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