How do you differentiate #y=(2x-5)^4(8x^2-5)^-3#?

Question

Christopher Agresta · Accepted Answer

#y'=(8(2x-5)^3(-4x^2+30x-5))/((8x^2-5)^4)#

You would apply the rule:

#y=f(x)g(x)->y'=color(red)(f'(x))g(x)+f(x)color(green)(g'(x))#

and the chain rule, since

#f(x)=(2x-5)^4 and g(x)=(8x^2-5)^-3#

Then:

#y'=color(red)(4(2x-5)^3*2)(8x^2-5)^-3+(2x-5)^4*color(green)((-3)(8x^2-5)^-4*16x)#

#=8(2x-5)^3/(8x^2-5)^3-48x(2x-5)^4/(8x^2-5)^4#

#=(8(8x^2-5)(2x-5)^3-48x(2x-5)^4)/((8x^2-5)^4)#

#=(8(2x-5)^3(8x^2-5-12x^2+30x))/((8x^2-5)^4)#

#=(8(2x-5)^3(-4x^2+30x-5))/((8x^2-5)^4)#

Samantha Adkison · Answer

undefined To differentiate the function y=(2x-5)^4(8x^2-5)^-3, you would use the product rule and the chain rule. The product rule states that if you have two functions, u(x) and v(x), their derivative is u'(x)v(x) + u(x)v'(x).

Let's denote u(x) = (2x - 5)^4 and v(x) = (8x^2 - 5)^-3. Then, u'(x) = 4(2x - 5)^3 * 2 and v'(x) = -3(8x^2 - 5)^-4 * 16x.

Now, applying the product rule:

y' = u'(x)v(x) + u(x)v'(x)
   = 4(2x - 5)^3 * 2 * (8x^2 - 5)^-3 + (2x - 5)^4 * (-3)(8x^2 - 5)^-4 * 16x

Thus, the derivative of y with respect to x is:

y' = 8(2x - 5)^3(8x^2 - 5)^-3 - 48x(2x - 5)^4(8x^2 - 5)^-4