Find f′ in terms of g′ f(x)=x2g(x)

Select one:

f′(x)=2xf′(x)+2xg′(x)

f′(x)=2xg′(x)

f′(x)=2x+g′(x)

f′(x)=x2g(x)+2x2g′(x)

f′(x)=2xg(x)+x2g′(x)

f(x)g(x) if you find the derivative of two products, you get f(x)g’(x)+f’(x)g(x). Let’s say x^2 = f(x).

(x^2)(g’(x))+2x(g(x)) so it would be your last option, 2x(g(x))+x^2(g’(x))

answer:   28.56 feet (approx)

step-by-step explanation:

since, triangle abc has measure of angle c equal to 55 degrees, measure of angle abc equal to 90 degrees, and length of bc equal to 20 feet.

we have to find out the measurement of segment ab.

therefore, tan 55° =

but bc = 20 feet.

⇒ tan 55 ° =  

⇒ ab = 20 × tan 55 °

⇒ ab = 20 × 1.42814800674

⇒ ab = 28.5629601348   ≈ 28.56 feet