Question

Find the derivative of the function

s(x)= e^4x-1 / x^3 - 1

Answer 1

\[= e ^{4x-1} \div x^3-1\]

Answer 2

Tried quotient rule already?

Answer 3

no i dont even know how to start

Answer 4

Ah, ok, well the quotient rule says when you're taking the derivative of a quotient like this one, then it's "low-d-high minus high-d-low, all over low-squared."

Answer 5

i.e. \(\large d/dx [ \frac{g(x)}{f(x)}] = \frac{f(x)g'(x)-g(x)f'(x)}{(f(x))^2}\)

Answer 6

ok. 4e^4x-1/(x^3-1)^2?

Answer 7

Takes a bit more than that.

Answer 8

\[4e ^{4x-1}/(x^3-1)^2\]

Answer 9

oh.

Answer 10

It's bottom times the derivative of the top minus top times the derivative of the bottom, all over bottom-squared.

Answer 11

whas the derivative of the top? 4x?

Answer 12

D[e^{4x-1}] = 4e^{4x-1}

Answer 13

Chain rule needed there.

Answer 14

so (4e^4x-1)(x^3-1)- (4e^4x-1)(x^3-1) / (x^3-1)^2 ?

Answer 15

That looks right. There is some simplifying you can do, but not much.

Answer 16

what would be simplfied? wouldnt the top just be canceled out?

Answer 17

Not everything would cancel out, no. There are a couple options, you can divide both terms in the numerator by the denominator, or you can factor the numerator.
You can continue to multiply things out, but there isn't much benefit to that.

Answer 18

ok so it would leave me with both 4e... equations after i cancel both the demoninator and numerator..

Answer 19

sorry im a little confused at this point

Answer 20

You might be better off leaving it in the form you have it, then.

My preferred way would be to leave the top and bottom factored like this:

\(\large \frac{e^{4x-1}(4x^3-3x^2-4)}{(x^3-1)^2}\)