Question

assume that ( f(x) ) and ( g(x) ) are differentiable at ( x ). find an expression for the derivative of ( y ).
( y = f(x)+6g(x)g(x) )
choose the correct answer below.
( \bigcirc ) a. ( f(x)+6g(x)g(x)+f^{prime}(x)+6g^{prime}(x)g^{prime}(x) )
( \bigcirc ) b. ( f(x)+6g^{prime}(x)g(x)+f^{prime}(x)+6g(x)g^{prime}(x) )
( \bigcirc ) c. ( f^{prime}(x)+6g^{prime}(x)g(x)+g^{prime}(x)f(x)+6g(x) )
( \bigcirc ) d. ( f^{prime}(x)+6g(x)g(x)+f(x)+6g^{prime}(x)g^{prime}(x) )

Explanation:

Step1: Apply the product rule

The product rule states that if \(y = u\cdot v\), then \(y^\prime=u^\prime v + uv^\prime\). Let \(u = f(x)+6g(x)\) and \(v = g(x)\).
First, find \(u^\prime\):
Using the sum rule \((a + b)^\prime=a^\prime + b^\prime\) and the constant - multiple rule \((cf(x))^\prime = cf^\prime(x)\), we have \(u^\prime=(f(x)+6g(x))^\prime=f^\prime(x)+6g^\prime(x)\) and \(v^\prime = g^\prime(x)\)

Step2: Substitute into the product rule

By the product rule \(y^\prime=u^\prime v+uv^\prime\), substituting \(u = f(x)+6g(x)\), \(u^\prime=f^\prime(x)+6g^\prime(x)\), \(v = g(x)\) and \(v^\prime = g^\prime(x)\) gives \(y^\prime=[f^\prime(x)+6g^\prime(x)]g(x)+[f(x)+6g(x)]g^\prime(x)\)

Answer:

C. \([f^{\prime}(x)+6g^{\prime}(x)]g(x)+g^{\prime}(x)[f(x)+6g(x)]\)