What is the Derivative of y=sin²x² ?

3 Answers
Apr 28, 2018

#(dy)/(dx)=4xsinx^2cosx^2=2xsin2 x^2#

Explanation:

We know that,

#color(red)((1)d/(dX)(X^n)=n*X^(n-1)#

#color(violet)((2)d/(dX)(sinX)=cosX#

#color(brown)((3)sin2theta=2sinthetacostheta#

Here,

#y=sin^2 x^2=(sinx^2)^2#

#"Using "color(blue)"Chain Rule",#step by step

#(dy)/(dx)=color(red)2(sinx^2)^color(red)1d/(dx)(sinx^2)...tocolor(red)(Apply(1)#
#color(white)((dy)/(dx))=2(sinx^2)color(violet)((cosx^2))d/(dx)(x^2)...tocolor(violet)(Apply(2)#
#color(white)((dy)/(dx))=2(sinx^2)(cosx^2)(2x)#
#(dy)/(dx)=4xsinx^2cosx^2#
#color(white)((dy)/(dx))=2x[color(brown)(2sinx^2cosx^2)]...tocolor(brown)(Apply(3)#
#color(white)((dy)/(dx))=2xsin2 x^2#

Apr 28, 2018

#dy/dx=2xsin2sin2x^2#

Explanation:

#"differentiate using the "color(blue)"chain rule"#

#"Given "y=f(g(x))" then"#

#dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"#

#y=sin^2x^2=(sinx^2)^2#

#rArrdy/dx=2sinx^2xxd/dx(sinx^2)#

#color(white)(rArrdy/dx)=2sinx^2xxcosx^2xxd/dx(x^2)#

#color(white)(rArrdy/dx)=4xsinx^2cosx^2=2xsin2x^2#

Apr 28, 2018

#dy/dx=2xsin(2x^2)#

Explanation:

#y=sin^2(x^2)# can be differentiated by the chain rule:

#(df(u))/dx=((df)/(du))*((du)/(dx))#

Where:
#f=u^2# and #u=sin(x^2)#

#(df)/(du)(u^2)=2u#

#(du)/dx(sin(x^2))# using the chain rule:
#f=sinu# and #u=x^2#

#d/(du)(sinu)d/dx(x^2)#:
#d/(du)(sinu)=cos(u)#
#d/dx(x^2)=2x#

Apply chain rule:

#(du)/dx=cos(u)*2x#
#(du)/dx=cos(x^2)*2x# (substituted #u# back in)

Therefore,

#dy/dx=(df)/(du)*(du)/(dx)#

#dy/dx=2ucos(x^2)*2x#
#dy/dx=2sin(x^2)cos(x^2)*2x#

Using following identity: #2cos(x)sin(x)=sin(2x)#

#dy/dx=2xsin(2x^2)#