How do you differentiate #f(x)=sqrt(sin^2x^2 - cos^3x)# using the chain rule?

1 Answer

Answer:

#color(red)(f' (x)=(4x*sin x^2*cos x^2+3*sin x*cos^2 x)/(2sqrt(sin^2 x^2-cos^3 x)))#

Explanation:

From #f(x)=sqrt(sin^2 x^2-cos^3 x)#
#f' (x)=d/dx(sqrt(sin^2 x^2-cos^3 x))#

Using #d/dx(sqrtu)=1/(2sqrtu)*d/dx(u)#

#f' (x)=d/dx(sqrt(sin^2 x^2-cos^3 x))#

#f' (x)=1/(2(sqrt(sin^2 x^2-cos^3 x)))*d/dx(sin^2 x^2-cos^3 x)#

#f' (x)=1/(2(sqrt(sin^2 x^2-cos^3 x)))*[2(sin x^2)^(2-1)d/dx(sin x^2)-3(cos x)^(3-1)d/dx(cos x)]#

#f' (x)=1/(2(sqrt(sin^2 x^2-cos^3 x)))*[2(sin x^2)(cos x^2)d/dx(x^2)-3(cos^2 x)(-sin x)]#

#f' (x)=1/(2(sqrt(sin^2 x^2-cos^3 x)))*[2(sin x^2)(cos x^2)(2x)-3(cos^2 x)(-sin x)]#

#color(red)(f' (x)=(4x*sin x^2*cos x^2+3*sin x*cos^2 x)/(2sqrt(sin^2 x^2-cos^3 x)))#

God bless....I hope the explanation is useful.