$g(x)=4x-\dfrac{1}{4x}=4x-(4x)^{-1}\\\\\\\\g'(x)=\left[4x-(4x)^{-1}\right]'=(4x)'-\left[(4x)^{-1}\right]'=4-[-1(4x)^{-2}]\cdot4\\\\=4+\dfrac{1}{(4x)^2}\cdot4=4+\dfrac{4}{4^2x^2}=\boxed{4+\dfrac{1}{4x^2}}\\\\Used:\\\\(f(x)-g(x))'=f'(x)-g'(x)\\\\.[f(g(x))]'=f'(g(x))\cdot g'(x)\\\\(x^n)'=nx^{n-1}$

8 0

3 years ago

How do I solve p(q-2)(r-3)

guajiro [1.7K]

Since you have parentheses, you will first multiply (q-2)*(r-3) which equals (qr+6). So now its simplified to p(qr+6) which equals pqr+p6.

5 0

3 years ago

What would you have to do to change 10 cubic feet into cubic inches?

Alina [70]

A. Divide by 46,656

B. Multiply by 1,728

C. Divide by 1,728

D. Multiply by 46,656

THE ANSWER IS B MULTIPLY BY 1,728

5 0

3 years ago