Question

question topic(s)/section(s): 2.9 the quotient rule
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$x$	0	3	5	9
$f(x)$	2	8	9	0
$g(x)$	5	4	-1	-4
$g(x)$	7	-5	-8	3

the functions $f$, $g$, $h$, and $k$ are twice - differentiable functions. let $h(x)=\frac{f(x)}{cos(x)}$. find $h(0)$.
0
3
2
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clear my selection

Explanation:

Step1: Recall the quotient - rule

The quotient - rule states that if $h(x)=\frac{u(x)}{v(x)}$, then $h^{\prime}(x)=\frac{u^{\prime}(x)v(x)-u(x)v^{\prime}(x)}{v^{2}(x)}$. Here, $u(x) = f(x)$ and $v(x)=\cos(x)$. So $h^{\prime}(x)=\frac{f^{\prime}(x)\cos(x)-f(x)(-\sin(x))}{\cos^{2}(x)}=\frac{f^{\prime}(x)\cos(x)+f(x)\sin(x)}{\cos^{2}(x)}$.

Step2: Evaluate $h^{\prime}(0)$

Substitute $x = 0$ into the formula for $h^{\prime}(x)$. We know that $\cos(0)=1$, $\sin(0)=0$, $f(0) = 3$, and $f^{\prime}(0)=2$.
$h^{\prime}(0)=\frac{f^{\prime}(0)\cos(0)+f(0)\sin(0)}{\cos^{2}(0)}$.
Substitute the values: $h^{\prime}(0)=\frac{2\times1 + 3\times0}{1^{2}}$.
$h^{\prime}(0)=\frac{2+0}{1}=2$.

Answer:

2