Question

Find the derivative of the following function:
y=(2x+1)^-3 (3x-7)^6

Answer 1

\[y= (2x + 1)^{-3} (3x-7)^{6}\]

Answer 2

I am also looking for a formula or something that could explain this derivative easily.

Answer 3

Product rule, power rule, chain rule.

Answer 4

Or quotient rule, if you're into that kind of thing

Answer 5

See, my teacher never thought us the rules, he just kinda.. did them haha

Answer 6

Ok

Answer 7

\[
\large \text{Differentiation Rules}
\]Let \(c\) be constant with respect to \(x\).
Let \(u\) and \(v\) vary with respect to \(x\).
\[
\begin{array}{r|ccc|l}
1&d(c) &=& 0&\text{Constants}\\ \
2&d(x) &=& dx&\text{The variable of differentiation}\\
3&d(|x|) &=& \frac{x}{|x|}~dx&\text{Absolute value function}\\
4&d(x^c) &=& cx^{c-1}~dx&\text{Power functions}\\
5&d(\sqrt x) &=& \frac{dx}{2\sqrt x}&\text{Square root function}\\
6&d(c^x) &=& c^{x}\ln(c)~dx&\text{Exponential functions}\\
7&d(e^x) &=& e^x~dx&\text{Exponential function}\\
8&d(\log_c(x)) &=& \frac{dx}{x\ln(c)}&\text{Logarithmic functions}\\
9&d(\ln(x)) &=& \frac{dx}{x}&\text{Natural logarithmic functions}\\
\hline
10&d(cu) &=& c~du&\text{Constant times a function}\\
11&d(u+v) &=& du+dv&\text{Sum of functions}\\
12&d(u\cdot v) &=& v~du+u~dv&\text{Product of functions}\\
13&d\left(\frac uv\right) &=& \frac{v~du-u~dv}{v^2}&\text{Quotient of functions}\\
14&d(u\circ v) &=& \frac{du}{dv}~dv&\text{Composition of functions}\\
\end{array}
\]

Answer 8

I just a little explanation of Why and How

Answer 9

Oh my.

Answer 10

Start with number \(13\), quotient rule \[
y=\frac{(3x-7)^6}{(2x+1)^3} = \frac uv
\]

Answer 11

We need the derivative of \(u\), in the numerator: \[
u = (3x-7)^6
\]

Answer 12

6(3x-7)^5

Answer 13

close, but not quite

Answer 14

We'll use the composition rule. \[
w=3x-7,\quad u = w^6
\]

Answer 15

So we think of \(u\) as a function of \(w\). We differentiate with respect to \(w\):\[
du =6w^5 ~dw
\]

Answer 16

Then we find \(dw\): \[
dw = \frac{dw}{dx}~dx = 3~dx
\]

Answer 17

We can plug our \(x\) parts back in: \[
du = 6w^5~dw = 6(3x-7)^5(3~dx) = 18(3x-7)^5~dx
\]

Answer 18

So, at least now we can say: \[
\frac{du}{dx} = 18(3x-7)^5
\]

Answer 19

Okay. Give me a minute to get this down. This is a little more complex then I was taught.

Answer 20

We need to find \(dv/dx\), which will be very similar.

Answer 21

\[
v=(2x+1)^3
\]

Answer 22

So it would be 6(2x+)^3?

Answer 23

(2x+1) sorry

Answer 24

Well, the \(3\) will go to \(2\)

Answer 25

shoot... that was a little mistake, that's what I meant.

Answer 26

Okay, assuming that you have the derivatives of \(u\) and \(v\)...\[
dy = \frac{v~du-u~dv}{v^2}
\]

Answer 27

\[\frac{ (2x+1) 18(3x-7)^{5} - (3x - 7) 6(2x+1)^{2} }{ (2x+1)^6 }\]

Answer 28

close, but your powers are off

Answer 29

Yes. I forgot to add the powers to my du. (2x+1)^3 and dv (3x -7)^6