assume that ( f(x) ) and ( g(x) ) are differentiable at ( x ). find an expression for the derivative of ( y…

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

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

Answer

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)])