Question

y" for y=(x^2+9)^4

Answer 1

use chain rule

Answer 2

I did use chain rule but got confused when doing the steps

Answer 3

Start from the outermost part of the function, the power to the fourth.

So we will say that \[y=z^4\]
Where \[z=x^2+9\]

So,\[y'=4z^3\]

Now we must deal with the inner part of the function, the x^2+9, which I called "z"
So,
\[z'=2x\]

Putting these together with the chain rule will yield
\[y'=4(x^2+9)^3(2x)\]or
\[y'=8x(x^2+9)^3\]

Now do it all again for y'' but this time we have to deal with the product rule.

So the product rule is\[(uv)'=uv'+u'v\]

Where u=8x and v=(x^2+9)^3

So you must find the derivative of both u and v.

\[u'=8\]

To find v' you must use the chain rule again.

This time the inner function is z=x^2+9

and the outer is z^3

So,\[v'=3(x^2+9)^2(2x)\]or\[v'=6x(x^2+9)^2\]

Put this all together and get
\[y''=(8x)(6x)(x^2+9)^2+8(x^2+9)^3\]

Simplify to get:
\[y''=48x^2(x^2+9)^2+8(x^2+9)^3\]

I hope this makes sense, the chain rule is very difficult to explain

Answer 4

Thank you so much that really helped a lot

Answer 5

From here:
\[ \frac{d}{dx}((x^2+9)^4) \]
Use the chain rule,
\[\frac{d}{dx}((x^2+9)^4) = \frac{ du^4}{ du} * \frac{du}{dx} \]
Let u = x^2+9
and remember that
\[\frac{du^4}{du} = 4 u^{4-1} = 4 u^3 \]
substitute back in, chain to the next shell layer
\[4 (x^2+9)^3 *\frac{d}{dx}(x^2+9) \]
Keep going....
\[4 (x^2+9)^3 * ( \frac{d}{dx}(x^2)+\frac{d}{dx}(9) ) \]
The derivative of 9 is zero and the derivative of x^2 is 2 x:
\[y' = 4(x^2+9)^3 (2x)\]

You do this process again to the result of the first derivative. You'll get something like this:

\[y′′=48x^2(x2+9)^2+8(x2+9)^3\]
Try it!

Remember with the chain rule it's you do a derivative to the outer shell function first, then to the inner until you get something like
\[ \frac{d}{dx} x = \frac{dx}{dx} = 1\]
or for a constant # 'k'
\[ \frac{d}{dx} k = 0\]
because the slope of a constant is zero