How do you find the derivative of #x*(sqrt(4-x^2))#?

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Answer 1 · 2017-03-23T04:44:08

#dy/dx=(-2x^2)/(2 sqrt(4-x^2))+sqrt(4-x^2)#

Before offering a solution to the problem, let me tell you an easy method of finding the derivative of #sqrtx#

You put the derivative of #x# in the numerator and multiply the given function with 2 and put it in the denominator.

The derivative of #sqrt x= 1/(2sqrtx)#

Use this simple technique in the given problem

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Now the problem

Given -

#y=x. (sqrt(4-x^2))#

#dy/dx=x.[(-2x)/(2 sqrt(4-x^2))]+sqrt(4-x^2).(1)#

#dy/dx=(-2x^2)/(2 sqrt(4-x^2))+sqrt(4-x^2)#

Answer 2 · 2017-03-23T04:44:49

#f'(x) = (4-2x^2)/sqrt(4-x^2)#

#f(x) = x*sqrt(4-x^2)#

#f'(x) = x*d/dx(4-x^2)^(1/2) + (4-x^2)^(1/2) * 1# [Product rule]

#= x*1/2*(4-x^2)^(-1/2)* (-2x) + (4-x^2)^(1/2) # [Chain rule]

#= (x *-2x)/(2sqrt(4-x^2)) + sqrt(4-x^2)#

#= sqrt(4-x^2) - x^2/sqrt(4-x^2)#

#= (4-x^2-x^2)/sqrt(4-x^2)#

#= (4-2x^2)/sqrt(4-x^2)#