Answer:
[tex]\displaystyle h'(0) = -5[/tex]
Step-by-step explanation:
We are given the function:
[tex]\displaystyle h(x) = 3f(x) -2g(x) - 5\cos x - 3[/tex]
And that f'(0) = 3 and g'(0) = 7.
And we want to determine the value of h'(0).
Find h'(x). We can take the derivative of both sides:
[tex]\displaystyle h'(x) = \frac{d}{dx}\left[ 3f(x) - 2g(x) - 5\cos x - 3\right][/tex]
Expand and simplify:
[tex]\begin{aligned}\displaystyle h'(x) & = \frac{d}{dx}\left[ 3f(x) - 2g(x) - 5\cos x - 3\right] \\ \\ & = \frac{d}{dx}[3f(x)] + \frac{d}{dx}[-2g(x)] + \frac{d}{dx}[-5\cos x] + \frac{d}{dx}[-3]\\ \\ & = 3f'(x) -2g'(x) +5\sin x\end{aligned}[/tex]
Therefore:
[tex]\displaystyle h'(0) = 3f'(0) -2 g'(0) + 5 \sin (0)[/tex]
Substitute and evaluate:
[tex]\displaystyle \begin{aligned} h'(0) & = 3(3) - 2(7) + 5(0) \\ \\ & = (9) - (14) + (0) \\ \\ & = -5\end{aligned}[/tex]
In conclusion:
[tex]\displaystyle h'(0) = -5[/tex]
