How do you find the derivative for #y= ((e^x)/(x^2))#?

Question

Emilia Aguilar · Accepted Answer

You use the quotient rule.

Notice that your function can be written as the quotient of two other functions, let's say #f(x)# and #g(x)#, which means that you can differentiate it by using the quotient rule. #color(blue)(y^' = (f^'(x) * g(x) - f(x) * g^'(x))/[g(x)]^2#, where #g(x)!=0# In your case, you have #y = f(x)/g(x)#, where #f(x) = e^x# and #g(x) = x^2# This means that your derivative will look like this #y^' = (d/dx(e^x) * x^2 - e^x * d/dx(x^2))/(x^2)^2# The derivative of #e^x# is always #d/dx(e^x) = e^x#, so you get #f^' = (e^x * x^2 - e^ * (2x))/x^4 = (x * e^x * (x - 2))/x^4# You can simplify this to be #f^' = (color(red)(cancel(color(black)(x))) * e^x (x-2))/x^(color(red)(cancel(color(black)(4)))color(blue)(3)) = color(green)((e^x(x-2))/x^3)#

Luke Alvarez · Answer

undefined To find the derivative of $ y = \frac{e^x}{x^2} $, you can use the quotient rule. The quotient rule states that for functions $ u(x) $ and $ v(x) $ where $ v(x) \neq 0 $, the derivative of $ \frac{u(x)}{v(x)} $ is given by:

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

Applying the quotient rule to $ y = \frac{e^x}{x^2} $, where $ u(x) = e^x $ and $ v(x) = x^2 $, we get:

$$ y' = \frac{(e^x)(2x) - (e^x)(2x)}{(x^2)^2} $$

$$ y' = \frac{2xe^x - 2xe^x}{x^4} $$

$$ y' = \frac{0}{x^4} $$

$$ y' = 0 $$

So, the derivative of $ y = \frac{e^x}{x^2} $ is $ y' = 0 $.