Question

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

\(5\)