Question

y=2^x.sin(tanx).dy/dx

Answer 1

Answer on Question #72508, Math / Differential Equations.

Task. $y = 2^x \cdot \sin(\tan x) \cdot dy / dx$

Solution.

1) We find $\frac{dy}{dx}$ where $y = 2^x \sin(\tan x)$.

So,

$$\frac{dy}{dx} = \left(2^x \sin(\tan x)\right)' = \left(2^x\right)' \sin(\tan x) + 2^x \left(\sin(\tan x)\right)' =$$ $$= 2^x \ln 2 \sin(\tan x) + 2^x \cos(\tan x)(\tan x)' = 2^x \ln 2 \sin(\tan x) + 2^x \cos(\tan x) \frac{1}{\cos^2 x}.$$

Answer: $\frac{dy}{dx} = 2^x \ln 2 \sin(\tan x) + 2^x \cos(\tan x) \frac{1}{\cos^2 x}$.

2) We find the general solution of the equation $y = 2^x \sin(\tan x) \cdot \frac{dy}{dx}$.

So,

$$\frac{dy}{y} = \frac{dx}{2^x \sin(\tan x)},$$ $$\int \frac{dy}{y} = \int \frac{dx}{2^x \sin(\tan x)},$$ $$\ln y = \int \frac{dx}{2^x \sin(\tan x)},$$ $$y = e^{\int \frac{dx}{2^x \sin(\tan x)}}.$$

Answer: $y = e^{\int \frac{dx}{2^x \sin(\tan x)}}$