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

Question

Charles Arocha · Accepted Answer

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

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!

Charles Arocha · Answer

undefined To differentiate $ y = \frac{\sin(x)}{x^2} $, you can use the quotient rule:

$$ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{vu' - uv'}{v^2} $$

Where $ u = \sin(x) $ and $ v = x^2 $.

$$ u' = \cos(x) \quad \text{and} \quad v' = 2x $$

Now, plug these values into the quotient rule:

$$ \frac{d}{dx} \left( \frac{\sin(x)}{x^2} \right) = \frac{x^2 \cos(x) - 2x \sin(x)}{x^4} $$

Simplify the expression:

$$ \frac{d}{dx} \left( \frac{\sin(x)}{x^2} \right) = \frac{x \cos(x) - 2 \sin(x)}{x^3} $$

So, the derivative of $ y = \frac{\sin(x)}{x^2} $ is $ \frac{x \cos(x) - 2 \sin(x)}{x^3} $.