Question

Can somebody check my answer?

f(x)=5/((3-0.5x)^6)
f'(x)= ?
FOR THE ANSWER LOOK BELOW!

Answer 1

\[f'(x)=5' * \frac{ 1 }{ (3-0.5x)^6 }'=5'*((3-0.5x)^{-6})'= (3-0.5x)^{-6}+3(3-0,5x)^{-7}\]

Answer 2

As a multiplied constant... the 5 stays and is multiplied by the -6 in the first step of using the power rule for derivatives... then you must find the derivative of the inside (by chain rule) and multiply it. Changing the power to -7 is correct.

Answer 3

I know, that I should use formula of
\[\left( \frac{ f(x) }{ f(x) } \right)'=\frac{ f'(x)*g(x)+g'(x)*f(x) }{ ^{g^2(x)} }\]

But isnt it correct as I did?

Answer 4

You don't have to use the quotient rule... there is a way much like your way... there are just a couple errors.

Answer 5

Ah, thank you. : )

Answer 6

Your first conversion is perfect (changing to the negative exponent).

Answer 7

Ultimately, I got a 15 in front.

Answer 8

Yup, I forget 5 : ) Thank you. I got

\[\frac{ (3-0.5x)^{6} +15(3x-0.5x)^{5}}{ (3-0.5x)^{12} }=\frac{ (3-0.5x)^{5}(18-0.5x) }{ (3-0.5x)^{12}}=simplify=\frac{ 18-0.5x }{(3-0.5x)^{7}}\]

Answer 9

I can't see the far right side, but I have issues with what I DO see. My first step of a quotient rule looks like...\[\frac{(3-0.5x)^{6}(0)-5(6)(3-0.5x)^{5}(-0.5)}{(3-0.5x)^{12}}\]With the quotient rule, that 5 does turn to zero and cancels out that first term leaving...\[\frac{15(3-0.5x)^{5}}{(3-0.5x)^{12}}=\frac{15}{(3-0.5x)^{7}}\]

Answer 10

With your other method...\[\frac{d}{dx}[5(3-0.5x)^{-6}]=5(-6)(3-0.5x)^{-7}(-0.5)\]Which simplifies to...\[\frac{15}{(3-0.5x)^{7}}\]

Answer 11

Ah. My bad. I always thought a'=1, where a is a number. Thank you a lot for your help ;)

Answer 12

You're welcome. :)