What is the derivative of $5^{tan x}$ $5^tanx$ ?

Question

2927 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-03T10:48:00

$\Rightarrow \frac{d y}{d x} = 5^{tan x} \cdot ln 5 \cdot {sec}^{2} x$ $=> (dy)/(dx) = 5^tanx * ln5 * sec^2 x$

We are trying to find $\frac{d y}{d x}$ $(dy)/(dx)$ when $y = 5^{tan x}$ $y = 5^tanx$

Taking natural logs on both sides...

$\Rightarrow ln y = {ln 5}^{tan x}$ $=> ln y = ln 5^tanx$

Using log laws:

$α log β \equiv {log β}^{α}$ $alpha log beta -= log beta ^ alpha$

$\Rightarrow ln y = ln 5 \cdot tan x$ $=> ln y =ln5 * tanx$

Differentiating implicitly:

$\Rightarrow \frac{1}{y} \cdot \frac{d y}{d x} = ln 5 \cdot {sec}^{2} x$ $=> 1/y * (dy)/(dx) = ln5 * sec^2 x$

$\Rightarrow \frac{d y}{d x} = y ln 5 \cdot {sec}^{2} x$ $=> (dy)/(dx) = y ln5 * sec^2 x$

$\Rightarrow \frac{d y}{d x} = 5^{tan x} \cdot ln 5 \cdot {sec}^{2} x$ $=> (dy)/(dx) = 5^tanx * ln5 * sec^2 x$

What is the derivative of 5^tanx5tanx5^tanx?