Question

What is the Derivative of y=sin²x² ?

Answer 1

`(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`

Answer 2

`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`

Answer 3

`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)`