How do you differentiate $y = \frac{e^{x}}{x^{7}}$ $y=e^x/x^7$ ?

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Answer 1 · 2016-07-30T14:58:19

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

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:

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

How do you differentiate y=e^x/x^7y=exx7y=e^x/x^7?