What’s 8.43 x 10^7 in standard form?

Ответ:

alannaamarriee

18.09.2022

$r'(1) = 60$ ====== $r(x) = f\big( g\big( h(x) \big) \big)$ to differentiate, we need to repeatedly use the chain rule. $\begin{aligned} r'(x)& = \frac{d}{dx} \left[f\big( g\big( h(x) \big) \big)\right] \end{aligned}$ the outside function is $f(x)$ ; the inside function is $g\big( h(x) \big)$ .the chain rule says to differentiate the outside function and leave the inside function inside. then multiply that by the derivative of the inside function. $\begin{aligned} r'(x) & = \frac{d}{dx} \left[f\big( g\big( h(x) \big) \big)\right] \\ & = f'\big( g\big( h(x) \big) \big) \cdot \frac{d}{dx} g\big( h(x) \big) \end{aligned}$ after that, repeat chain rule for $(d/dx)\ g\big( h(x) \big)$ , using $g(x)$ as the outside function, $h(x)$ as inside. $\begin{aligned} r'(x) & = f'\big( g\big( h(x) \big) \big) \cdot g'\big( h(x) \big) \cdot \frac{d}{dx} h(x)\\ r'(x) & = f'\big( g\big( h(x) \big) \big) \cdot g'\big( h(x) \big) \cdot h'(x) \end{aligned}$ since we are looking for r'(1): $\begin{aligned} r'(1) & = f'\big( g\big( h(1) \big) \big) \cdot g'\big( h(1) \big) \cdot h'(1) \\ & = f'\big( g\big( \boxed{h(1)} \big) \big) \cdot g'\big( h(1) \big) \cdot h'(1) \end{aligned}$ we are given $h(1) = 5$ $r'(1) = f'\big( g\big( 5 \big) \big) \cdot g'\big( h(1) \big) \cdot h'(1)$ we are given $g(5) = 3$ $r'(1) = f'( 3)\cdot g'\big( h(1) \big) \cdot h'(1)$ we are given $f'( 3) = 5$ $r'(1) = 5\cdot g'\big( h(1) \big) \cdot h'(1)$ again, $h(1) = 5$ $r'(1) = 5\cdot g'\big( 5\big) \cdot h'(1)$ given $g'\big( 5\big) = 3$ $r'(1) = 5\cdot 3 \cdot h'(1)$ given $h'(1) = 4$ $\begin{aligned} r'(1) & = 5\cdot 3 \cdot 4 \\ & = 15 \cdot 4 \\ & = 60 \end{aligned}$

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