Question

Finding Derivatives Implicity In Exercise,Find dy/dx implicity. xey - 10x + 3y = 0

Answer 1

Answer:

$\dfrac{dy}{dx}=\dfrac{10-e^y}{(xe^y+3)}$

Step-by-step explanation:

Given:

The implicit equation is given as:

$xe^y-10x+3y=0$

In implicit differentiation, we treat 'y' as a function of 'x' and differentiate both sides of the equation with respect to 'x' and then collect all the $\frac{dy}{dx}$ together and finally solve for $\frac{dy}{dx}$.

So, differentiating both sides of the above equation with respect to 'x'. This gives,

$\frac{d}{dx}(xe^y-10x+3y)=\frac{d}{dx}(0)\\\\\frac{d}{dx}(xe^y)+\frac{d}{dx}(-10x)+\frac{d}{dx}(3y)=0\\\\\textrm{Using product rule, (uv)' = uv' + vu'}\\\\(x\cdot e^y\cdot \frac{dy}{dx}+e^y\cdot 1)-10\cdot1+3\frac{dy}{dx}=0\\\\\frac{dy}{dx}(xe^y)+e^y-10+\frac{dy}{dx}(3)=0\\\\\textrm{Grouping}\ \frac{dy}{dx}\textrm{ terms together}\\\\\frac{dy}{dx}(xe^y+3)+e^y-10=0\\\\\frac{dy}{dx}(xe^y+3)=10-e^y\\\\\dfrac{dy}{dx}=\dfrac{10-e^y}{(xe^y+3)}$

Therefore, the derivative $\frac{dy}{dx}$ implicitly is:

$\dfrac{dy}{dx}=\dfrac{10-e^y}{(xe^y+3)}$