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

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Answer 1 · 2018-04-03T23:44:56

$ln49[7^[2x]]$

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

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

Therefore, $d/dx7^[2x=$ $d/[dx]$ $[e^[2xln7]]$

$d/dx[e^[2xln7]]$ = $[e^[2xln7]d/dx[2xln7]]$ and since $ln7$ is a constant,

$d/dx[2xln7]$ = $2ln7$ ..... So, $d/dx7^[2x$ = $2ln7[e^[2xln7]]$ ....... $[2]$

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

$d/dx 7^[2x$ = $2ln7[7^[2x]]$ = $ln49[7^[2x]]$ . Hope this was helpful.

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