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# Help with chain rule?

compute the derivative using the chain rule

(((4x^2)+2)^7)(((3x^3)+5x)^8)

i could have sworn it'd be (7(4x^2+2)^6)(8x)(8(3x^3+5x)^7)(9x^2+5), but it is not. How do I do it. Thanks you in advance

### 6 Answers

- 1 decade agoFavorite Answer
(4x² + 2)⁷(3x³ + 5x)⁸

Use the product rule first. Let f(x) = (4x² + 2)⁷ and g(x) = (3x³ + 5x)⁸.

Then the derivative of f(x)g(x) = f '(x)g(x) + f(x)g '(x). The only quantities we don't have are f '(x) and g '(x).

f(x) = (4x² + 2)⁷

f '(x) = 7(4x² + 2)⁶ * (d/dx)(4x² + 2) by the chain rule

= 7(4x² + 2)⁶ * 8x

= 56x(4x² + 2)⁶

g(x) = (3x³ + 5x)⁸

g '(x) = 8(3x³ + 5x)⁷ * (d/dx)(3x³ + 5x) by the chain rule

= 8(3x³ + 5x)⁷ * (9x² + 5)

Simply substitute all these expressions for f '(x)g(x) + f(x)g '(x)

= 56x(4x² + 2)⁶(3x³ + 5x)⁸ + 8(4x² + 2)⁷(3x³ + 5x)⁷(9x² + 5)

There's nothing you can really do to simplify that, so it's the final answer.

Edit: Beware. Every answer except for mine and δoτ's answer are incorrect.

- PolyhymnioLv 71 decade ago
You have to use the product rule first, then the chain rule for the two derivatives used in the product rule.

Let f(x) = (4x^2 + 2)^7

g(x) = (3x^3 + 5x)^8

f'(x) = 7(4x^2 + 2)^6(8x) = 56x(4x^2 + 2)^6

g'(x) = 8(3x^3 + 5x)^7(9x^2 + 5) = (72x^2 + 80)(3x^3 + 5)^7

(fg)' = f'g + fg' =

56x(4x^2 + 2)^6(3x^3 + 5x)^8 + (72x^2 + 80)(3x^3 + 5)^7(4x^2 + 2)^7

- Anonymous1 decade ago
You have to use the chain rule many times, but you also have to use the product rule once and unless I missed something, your answer seems to indicate that you did not use the product rule. I have written the original product with a * in it just below.

(((4x^2)+2)^7) * (((3x^3)+5x)^8)

First ((4x^2) + 2)^7

Let u = (4x^2) + 2

Then du/dx = 8x

Let y = u^7

Then dy/du = 7u^6 and since u = ((4x^2) + 2) it is 7 * ((4x^2) + 2)^6

So dy/dx = dy/du * du/dx = 7 * ((4x^2) + 2)^6 * 8x = 56x * ((4x^2) + 2)^6

---------------------

Now for (((3x^3)+5x)^8)

Let v = ((3x^3) + 5x)

Then dv/dx = 9x^2 + 5

Let y = v^8

Then dy/dv = 8v^7 and since v = ((3x^3) + 5x) it is also equal to:

(((8 * (((3x^3) + 5x))^7

so dy/dx = dy/du * du/dx = (8 * (((3x^3) + 5x))^7 ) * ((9x^2) + 5)

--------------

So wrapping this whole thing together, the entire derivative is:

We need: u * dv/dx + v * du/dx which is:

{[u = (4x^2) + 2] * [(8 * (((3x^3) + 5x))^7 ) * ((9x^2) + 5)]} +

{[((3x^3) + 5x)] * [56x * ((4x^2) + 2)^6]}

.

- δοτζοLv 71 decade ago
(4x²+2)⁷(3x³+5x)⁸

You did the individual derivatives correctly, but the product rule wasn't used, and it has to be.

(4x²+2)⁷*(8(3x³+5x)⁷(9x²+5)) + (3x³+5x)⁸*(7(4x²+2)⁶(8x))

8(4x²+2)⁶(3x³+5x)⁷ * [(4x²+5)(9x²+5) + 7x(3x³+5x)]

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- HemantLv 71 decade ago
Use lagarithms : Dx ( ln u ) = ( 1 / u ). du/dx

...............................................................................

Taking natural logs of both sides of the given eq,

ln y = 7 ln ( 4 x^2 + 2 ) + 8 ln ( 3 x^3 + 5x )

................................................................................

Diff....w.r.t. x,

( 1/y). dy/dx

= 7 [ 1/ ( 4x^2 + 2 ) ]. Dx( 4x^2 + 2 ) + 8 [ 1/ (3x^3 + 5x ) ]. Dx( 3x^3 + 5x )

dy/dx = y { 7 [ 8x / ( 4x^2 + 2x) ] + 8 [ ( 9x^2 + 5 ) / ( 3x^3 + 5x ) ] ]

........= y { [ 28 / ( 2x^2 + 1 ) ] + [ 8 ( 9x^2 + 5 ) / ( 3x^3 + 5x ) ] }

.......= ....you may plug in for y here.......................Ans.

........................................................................................

Happy To Help !