Anonymous

# 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

Relevance

(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.

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

• Anonymous

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]}

.

(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)]

• Hemant
Lv 7

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 !