Differentiate (x^2-4)(x+4)^7?

The answer is as follows:

y = (x² - 4)(x + 4)^7

This is a combination of the product rule and the chain rule.

Let's start breaking things up.

if: u = x² - 4  and v = (x + 4)^7

then:  y = uv

The product rule is:

y' = u'v + uv'

u' is easy enough to find, but v' needs the chain rule, so let's do that first:

If I say : w = x + 4 then v = w^7

Find the derivatives of both, then use the chain rule:

w = x + 4 and v = w^7
dw/dx = 1 and dv/dw = 7w^6

chain rule:

dv/dx = dv/dw * dw/dx
dv/dx = 7w^6 * 1
dv/dx = 7w^6

Now substitute the expression for "w" to get:

dv/dx = 7(x + 4)^6

That's our v':

v' = 7(x + 4)^6

Now we get our u', also reminding what u and v are:

u = x² - 4 and v = (x + 4)^7
u' = 2x and v = 7(x + 4)^6

Now we can substitute these into the product rule:

y' = u'v + uv'
y' = 2x(x + 4)^7 + (x² - 4) * 7(x + 4)^6

To try to simplify this a little, let's pull out one of the (x + 4)s from the (x + 4)^7 to leave (x + 4)^6, then simplify what's left:

y' = 2x(x + 4)(x + 4)^6 + 7(x² - 4)(x + 4)^6
y' = (2x² + 8x)(x + 4)^6 + (7x² - 28)(x + 4)^6

Now factor out (x + 4)^6 and simplify what's left:

y' = (2x² + 8x + 7x² - 28)(x + 4)^6
y' = (9x² + 8x - 28)(x + 4)^6

f(x) = (x^2-4)(x+4)^7. f'(x) = 2x*(x+4)^7+7(x^2-4)(x+4)^6= (x+4)^6(7x^2-28;
+2x^2+8x) = (9x^2+8x-28)(x+4)^6.