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)