Question

8. If sin(x + y) = log (x + y), then dy is
dx
equal to​

Answer 1

Answer:

first open the formulas od sin(x+y) and log(x+y) and took the y terms on same side to get its derivative.

Answer 2

$\dfrac{dy}{dx}=\dfrac{(x+y)\cos(x+y)+1}{x+y)\cos(x+y)-1}$

Step-by-step explanation:

sin(x + y) = log (x + y)

differentiate w.r.t x

$\cos(x+y)(1+\dfrac{dy}{dx})=\dfrac{1}{x+y}\cdot(1+\dfrac{dy}{dx})$

$\cos(x+y)+\cos(x+y)\dfrac{dy}{dx}=\dfrac{1}{x+y}+\dfrac{1}{x+y}\dfrac{dy}{dx}$

$\cos(x+y)\dfrac{dy}{dx}-\dfrac{1}{x+y}\dfrac{dy}{dx}=\cos(x+y)+\dfrac{1}{x+y}$

$\dfrac{dy}{dx}(\cos(x+y)-\dfrac{1}{x+y})=\cos(x+y)+\dfrac{1}{x+y}$

$\dfrac{dy}{dx}=\dfrac{\cos(x+y)+\dfrac{1}{x+y}}{\cos(x+y)-\dfrac{1}{x+y}}$

$\dfrac{dy}{dx}=\dfrac{(x+y)\cos(x+y)+1}{x+y)\cos(x+y)-1}$

#BAL

#ANSWERWITHQUALITY