Can someone differentiate this using product rule: (2x^2-5)^7?

You would only use product rule for equations that are being multiplied.

For this, there is a power of an expression, so you need to use the General Power Rule, a branch of the Chain Rule.

it states:

for any equation y = [u(x)]^n

then f'(u^n) = n(u)^(n-1) * u'

You are given

(2x^2-5)^7

So, the outer layer of the expression is [u(x)]^7

The inner layer [u(x)] is 2x^2-5

So, plugging into the formula, you get: 7(2x^2-5)^(6) * 4x

then go from there..

Easier to use the chain rule

dy/dx = 7(2x^2-5)^6 * 4x =

28x(2x^2-5)^6

It's a lot of work with the product rule but it can be done:

F(x)=(2x^2-5)^7=(2x^2-5)(2x^2-5)(2x^2-5)(2x^2-5)(2x^2-5)(2x^2-5)(2x^2-5).

product rule of the functions a to g =>

d/dx(a*b*c*d*e*f*g)=da/dx * bcdefg + db/dx * acdefg + dc/dx * abdefg + dd/dx abcefg ..... etc

since a=b=c=d=e=f=g=(2x^2-5)

F'(x)=((2x^2-5)^6)*(da/dx+db/dx+dc/dx+ ... dg/dx)

F'(x)=((2x^2-5)^6)*(28x)