If h(x) = sqrt{4f(x) + 3g(x)},f(1) = 4,g(1) = | vedclass

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Question

(B) Given $h(x) = \sqrt{4f(x) + 3g(x)}$.
At $x = 1$,$h(1) = \sqrt{4f(1) + 3g(1)} = \sqrt{4(4) + 3(3)} = \sqrt{16 + 9} = \sqrt{25} = 5$.
Now,differen

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$\frac{5}{2}$

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(B) Given $h(x) = \sqrt{4f(x) + 3g(x)}$.
At $x = 1$,$h(1) = \sqrt{4f(1) + 3g(1)} = \sqrt{4(4) + 3(3)} = \sqrt{16 + 9} = \sqrt{25} = 5$.
Now,differentiate $h(x)$ with respect to $x$ using the chain rule:
$h'(x) = \frac{1}{2\sqrt{4f(x) + 3g(x)}} \cdot \frac{d}{dx}(4f(x) + 3g(x))$
$h'(x) = \frac{4f'(x) + 3g'(x)}{2\sqrt{4f(x) + 3g(x)}}$
Substitute $x = 1$ into the derivative:
$h'(1) = \frac{4f'(1) + 3g'(1)}{2\sqrt{4f(1) + 3g(1)}}$
Given $f'(1) = 4$ and $g'(1) = 3$:
$h'(1) = \frac{4(4) + 3(3)}{2(5)} = \frac{16 + 9}{10} = \frac{25}{10} = \frac{5}{2}$.

If $h(x) = \sqrt{4f(x) + 3g(x)}$,$f(1) = 4$,$g(1) = 3$,$f'(1) = 4$,and $g'(1) = 3$,then find $h'(1)$.

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