suppose that ( h(x)=4 f(x)+3 g(x), f^{prime}(1)=2 ) and ( g^{prime}(1)=-1 ).\nfind ( h^{prime}(1) ).

Explanation:

Step1: Differentiate ( h(x) )

Using the sum rule and constant - multiple rule of differentiation. If ( h(x)=4f(x)+3g(x) ), then ( h^{\prime}(x)=\frac{d}{dx}(4f(x))+\frac{d}{dx}(3g(x)) ). By the constant - multiple rule (\frac{d}{dx}(cf(x)) = c\frac{d}{dx}(f(x))) (where (c) is a constant), we have ( h^{\prime}(x)=4f^{\prime}(x)+3g^{\prime}(x)).

Step2: Evaluate ( h^{\prime}(x) ) at ( x = 1 )

Substitute (x = 1) into (h^{\prime}(x)). We know that (f^{\prime}(1)=2) and (g^{\prime}(1)=-1). So (h^{\prime}(1)=4f^{\prime}(1)+3g^{\prime}(1)). Substitute the values: (h^{\prime}(1)=4\times2+3\times(-1)). First, calculate (4\times2 = 8) and (3\times(-1)=-3). Then (h^{\prime}(1)=8 - 3).

Answer:

(5)