Question

How do you differentiate `y=sinx/x^2`?

Answer 1

`dy/dx=(cos x)/(x^2)-(2sinx)/x^3`

Explanation:

We have

`y=sin x/x^2`

We can differentiate it by two methods:

Method 1:

We will use the quotient rule, there is a way I like to remember it:

Denominator same, differentiation of numerator. Minus numerator same, differentiation of denominator whole divided by denominator squared

Differentiating with respect to x:

`dy/dx=(x^2*cos x-sinx*2x)/(x^2)^2`

`dy/dx=(x^2*cos x-sinx*2x)/x^4`

`dy/dx=(cancel(x^2)*cos x)/(x^cancel(4))-(sinx*2cancel(x))/x^cancel(4)`

`dy/dx=(cos x)/(x^2)-(2sinx)/x^3`

Method 2:

Simplifying:

`y=sinx*x^-2`

We can now apply the product rule, there is a way I like to remember it:

First function same, differentiation of second function. Plus second function same, differentiation of first function

Differentiating with respect to x:

`dy/dx=sinx*(-2)x^-3+x^-2*cos x`

`dy/dx=-2sinx*x^-3+x^-2*cos x`

Rearranging the terms, we get:

`dy/dx=x^-2*cos x-2sinx*x^-3`

`dy/dx=(cos x)/(x^2)-(2sinx)/(x^3)`

I hope it helps!