How do you differentiate #y=e^x/x^7#?

Question

Christopher Agresta · Accepted Answer

#y^'=(e^x(x-7))/x^8#

The easiest way, for me, is to first write this not as a quotient:

#y=e^x/x^7=e^x x^-7#

From here, use the product rule, which states that if #y=f(x)g(x)#, then #y^'=f^'(x)g(x)+f(x)g^'(x)#.

So here, we see that #f(x)=e^x#, so #f^'(x)=e^x# as well, and #g(x)=x^-7#, so #g^'(x)=-7x^-8#.

Thus:

#y^'=e^x x^-7+e^x(-7x^-8)#

Simplifying:

#y^'=e^x/x^7-(7e^x)/x^8#

Common denominator:

#y^'=(xe^x-7e^x)/x^8#

#y^'=(e^x(x-7))/x^8#

Note that this can also be done with the quotient rule, which states that if #y=f(x)/g(x)# then #y^'=(f^'(x)g(x)-f(x)g^'(x))/(g(x))^2#.

So, in this case #f(x)=e^x# so again #f^'(x)=e^x#, but #g(x)=x^7# so #g^'(x)=7x^6#.

Thus:

#y^'=(e^x x^7-e^x(7x^6))/(x^7)^2=(e^x x^6(x-7))/x^14=(e^x(x-7))/x^8#

Emma Aldridge · Answer

undefined To differentiate $ y = \frac{e^x}{x^7} $, you can use the quotient rule. The quotient rule states that for functions $ u $ and $ v $, if $ y = \frac{u}{v} $, then the derivative $ \frac{dy}{dx} $ is given by:

$$ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} $$

In this case, $ u = e^x $ and $ v = x^7 $. So, we need to find the derivatives of $ u $ and $ v $, which are $ \frac{du}{dx} = e^x $ and $ \frac{dv}{dx} = 7x^6 $, respectively.

Applying the quotient rule, we get:

$$ \frac{dy}{dx} = \frac{x^7(e^x) - e^x(7x^6)}{(x^7)^2} $$

Simplify this expression to get the final result.