What is the derivative of # x * ((4-x^2)^(1/2))#?

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Answer 1 · 2016-10-24T08:03:24

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

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

I'll differentiate using the product and chain rules.

Product rule
#y=f(x)*g(x)#

#y'=f'(x)g(x) +g'(x)f(x)#

#y=f(g(x))#

#y' = f'(g(x))*g'(x)#

Differentiate each part
#f(x)=x#

#f'(x)=1#

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

( #y=f(g(x)) => y' = f'(g(x))*g'(x)#)

#g'(x)=1/2(4-x^-2)^(-1/2)(-2x)#

so using the product rule

#y'=f'(x)g(x) +g'(x)f(x)#

#y'=sqrt(4-x^2) +(1/2(4-x^-2)^(-1/2)(-2x))(x)#

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

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

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

which is the derivative of the function.