Question

Differentiate the following function.

y=[(x-2)(9x+4)]^3

Answer 1

Here you use the chain rule. Have you learned that yet?

Answer 2

d
dxf(g(x)) = f
0
(g(x))g
0
(x)

Answer 3

not yet, just the product and quotient rule

Answer 4

use that rule dood

Answer 5

compute whats in the brackets and then simplify until you have a regular polynomial

Answer 6

\[6(x-2)^{2}(9x-7)(9x+4)^{2}\]

Answer 7

is that the differentiation of the function?

Answer 8

simplify as much as possible

Answer 9

Use the chain rule! It will take way too long using the product rule.

Answer 10

I haven't learned the chain rule yet.

Answer 11

\(\large\color{black}{ \displaystyle y=\left[(x-2)(9x+4) \right]^3 }\)
when you take the derivative (dy/dx) (of function y, with respect to x) on both sides,

the left side is just dy/dx denoting that it (what is on the right side) is the 1st derivative of the function.

And the right side is as follows....

\(\large\color{black}{ \displaystyle \frac{ dy}{dx}=3\times \left[(x-2)(9x+4) \color{white}{\int}\right]^{3-1} \times \left[ \frac{ d}{dx} \left(~(x-2)(9x+4)~\right)\right]}\)

so we are applying the power rule to the entire argument, and if the derivative of the inner argument is not 1, then we apply the chain rule.

Answer 12

\(\large\color{black}{ \displaystyle \frac{ dy}{dx}=3\times \left[(x-2)(9x+4) \color{white}{\int}\right]^{3-1} \times \left[ \frac{ d}{dx} \left(~(x-2)(9x+4)~\right)\right]}\)

what will this be?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Well:
\(\large\color{black}{ \displaystyle \frac{ d}{dx} \left(~(x-2)(9x+4)~\right)}\)

product rule, (we get )

\(\large\color{black}{ \displaystyle (9x+4)\frac{ d}{dx} (x-2)~+~(x-2)\frac{ d}{dx} (9x+4)}\)

\(\large\color{black}{ \displaystyle (9x+4)1~+~(x-2)9}\)
\(\large\color{black}{ \displaystyle 9x+4~+~9x-18}\)
\(\large\color{black}{ \displaystyle 18x-14}\)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
So,

\(\large\color{black}{ \displaystyle \frac{ dy}{dx}=3\times \left[(x-2)(9x+4) \color{white}{\int}\right]^{3-1} \times \left[ \frac{ d}{dx} \left(~(x-2)(9x+4)~\right)\right]}\)

\(\large\color{black}{ \displaystyle \frac{ dy}{dx}=3\times \left[(x-2)(9x+4) \color{white}{\int}\right]^{3-1} \times \left[ 18x-14\right]}\)

\(\large\color{blue}{ \displaystyle \frac{ dy}{dx}=3\times \left[(x-2)(9x+4) \color{white}{\int}\right]^{2} \times \left[ 18x-14\right]}\)

this in blue is the derivative

Answer 13

It will be easier if you expand the product inside the brackets to yield quadratic, the follow the exponent and chain rules.

Answer 14

you can use logarithmic differentiation as well:)

Answer 15

it will be quite well coming out