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

1 Answer
Mar 15, 2017

#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!