What is the derivative of y = sqrt (cos x) / x^2* sin x?  

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to find the derivative of y = sqrt (cos x) / x^2* sin x

Now an easy way to find the derivative here is to use the logarithm.

take the natural logarithm of both the sides:

ln y = ln [sqrt (cos x) / x^2* sin x]

=> ln y = ln [sqrt (cos x)] – ln x^2 – ln sin x

=> ln y = (1/2) ln cos x – 2 ln x – ln sin x

Take the derivate of both the sides with respect to x

=> (1/y) dy/dx = (1/2) (–sin x)/ (cos x) – 2/x – cos x/ sin x

dy/dx = y*[(1/2) (–sin x)/ (cos x) – 2/x – cos x/ sin x]

dy/dx = y*[(-1/2) (tan x) – 2/x – cot x]

Therefore the required derivative of y = sqrt (cos x) / x^2* sin x is y*[(-1/2) (tan x) – 2/x – cot x]

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll aply the quotient rule and the chain rule:

f'(x) = y = [sqrt (cos x)]'*(x^2* sin x) - sqrt (cos x)*(x^2* sin x)'/(x^2* sin x)^2

f'(x) = -x^2*(sin x)^2/2sqrt(cos x) - sqrt (cos x)*(2xsin x + x^2*cos x)/(x^2* sin x)^2

f'(x) = [-x^2*(sin x)^2 - 4x*sin x*cos x - 2x^2*(cos x)^2]/(x^2* sin x)^2

f'(x) = {-x^2[(sin x)^2 + (cos x)^2] - xcos x(4 sin x + x*cos x)}/(x^2* sin x)^2

f'(x) = -[x^2 + xcos x(4 sin x + x*cos x)]/(x^2* sin x)^2

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