How do you find the derivative of $f (x) = 7^{2 x}$ $f(x)=7^(2x)$ ?

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How do you find the derivative of $f (x) = 7^{2 x}$ $f(x)=7^(2x)$ ?

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Answer 1 · 2018-04-03T23:44:56

$ln 49 [7^{2 x}]$ $ln49[7^[2x]]$

Explanation:

By the theory of logs $7^{2 x}$ $7^[2x$ can be written as $e^{2 x ln 7}$ $e^[2xln7]$ ,i.e,

$7^{2 x} = e^{2 x ln 7}$ $7^[2x]=e^[2xln7]$ ..... $[1]$ $[1]$

Therefore, $\frac{d}{d x} 7^{2 x =}$ $d/dx7^[2x=$ $\frac{d}{d x}$ $d/[dx]$ $[e^{2 x ln 7}]$ $[e^[2xln7]]$

$\frac{d}{d x} [e^{2 x ln 7}]$ $d/dx[e^[2xln7]]$ = $[e^{2 x ln 7} \frac{d}{d x} [2 x ln 7]]$ $[e^[2xln7]d/dx[2xln7]]$ and since $ln 7$ $ln7$ is a constant,

$\frac{d}{d x} [2 x ln 7]$ $d/dx[2xln7]$ = $2 ln 7$ $2ln7$ ..... So, $\frac{d}{d x} 7^{2 x}$ $d/dx7^[2x$ = $2 ln 7 [e^{2 x ln 7}]$ $2ln7[e^[2xln7]]$ ....... $[2]$ $[2]$

From ..... $[1]$ $[1]$ , $e^{2 x ln 7}$ $e^[2xln7$ = $7^{2 x}$ $7^[2x$ so substituting in 2,

$\frac{d}{d x} 7^{2 x}$ $d/dx 7^[2x$ = $2 ln 7 [7^{2 x}]$ $2ln7[7^[2x]]$ = $ln 49 [7^{2 x}]$ $ln49[7^[2x]]$ . Hope this was helpful.

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How do you find the derivative of $f (x) = 7^{2 x}$ $f(x)=7^(2x)$ ?

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Explanation:

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