How do you differentiate #f(x)= (2x^2-5)(4x^2+5) # using the product rule?

1 Answer
Dec 8, 2016

Answer:

# f'(x) = (4x)(8x^2-5) #

Explanation:

If you are studying maths, then you should learn the Product Rule for Differentiation, and practice how to use it:

# d/dx(uv)=u(dv)/dx+(du)/dxv #, or, # (uv)' = (du)v + u(dv) #

I was taught to remember the rule in words; "The first times the derivative of the second plus the derivative of the first times the second ".

So with # f(x) = (2x^2-5)(4x^2+5) # we have;

# { ("Let "u = 2x^2-5, => , (du)/dx =4x), ("And "v = 4x^2+5, =>, (dv)/dx = 8x ) :}#

# \ \ \ \ \ d/dx(uv) = u(dv)/dx + (du)/dxv #
# :. d/dx(uv) = (2x^2-5)(8x) + (4x)(4x^2+5) #
# :. d/dx(uv) = (4x){2(2x^2-5) + (4x^2+5)} #
# :. d/dx(uv) = (4x)(4x^2-10+4x^2+5) #
# :. d/dx(uv) = (4x)(8x^2-5) #