Solveeit Logo
Login

Question

Question: How do you use implicit differentiation to find \(\dfrac{dy}{dx}\) given \(x{{e}^{y}}-y=5\) ?...

How do you use implicit differentiation to find dydx\dfrac{dy}{dx} given xey−y=5x{{e}^{y}}-y=5 ?

Explanation

Solution

In this question, we have to find the differentiation of y with respect to x. therefore, we will use the differentiation formula to get the result for the solution. So, we start solving this problem by applying the product rule (u.v)′=u′v+uv′(u.v{)}'={u}'v+u{v}' on the left-hand side of the equation. Then, we will solve the differentiation of the given equation with respect to x. Then, we will subtract ey{{e}^{y}} on both sides of the equation and take dydx\dfrac{dy}{dx} common from the above equation. Thus, we will divide (xey−1)(x{{e}^{y}}-1) on both sides of the equation, to get the required solution for the problem.

Complete step-by-step solution:
According to the question, we have to find the differentiation of an equation.
Thus, we will apply the implicit differentiation to get the solution.
The equation given to us is xey−y=5x{{e}^{y}}-y=5 ---------- (1)
So, we will first apply the product rule (u.v)′=u′v+uv′(u.v{)}'={u}'v+u{v}' on the left-hand side of the equation (1) and further differentiate the equation with respect to x, we get
(x)′ey+x(ey)′−dydx=0(x{)}'{{e}^{y}}+x{{\left( {{e}^{y}} \right)}^{\prime }}-\dfrac{dy}{dx}=0
Therefore, differentiating y with respect to x, we get
1.ey+x(ey).dydx−dydx=01.{{e}^{y}}+x\left( {{e}^{y}} \right).\dfrac{dy}{dx}-\dfrac{dy}{dx}=0
On further simplification, we get
ey+xeydydx−dydx=0{{e}^{y}}+x{{e}^{y}}\dfrac{dy}{dx}-\dfrac{dy}{dx}=0
Now, we will subtract ey{{e}^{y}} on both sides in the above equation, we get
ey+xeydydx−dydx−ey=0−ey{{e}^{y}}+x{{e}^{y}}\dfrac{dy}{dx}-\dfrac{dy}{dx}-{{e}^{y}}=0-{{e}^{y}}
As we know, the same terms with opposite signs cancel out each other, therefore we get
xeydydx−dydx=−eyx{{e}^{y}}\dfrac{dy}{dx}-\dfrac{dy}{dx}=-{{e}^{y}}
Now, take dydx\dfrac{dy}{dx} common from the above equation, we get
dydx(xey−1)=−ey\dfrac{dy}{dx}(x{{e}^{y}}-1)=-{{e}^{y}}
Now, we will divide (xey−1)(x{{e}^{y}}-1) on both sides of the equation, we get
dydx(xey−1)(xey−1)=−ey(xey−1)\dfrac{dy}{dx}\dfrac{(x{{e}^{y}}-1)}{(x{{e}^{y}}-1)}=\dfrac{-{{e}^{y}}}{(x{{e}^{y}}-1)}
As we know, the same terms with the same signs cancel out with quotient 1, therefore we get
dydx=−ey(xey−1)\dfrac{dy}{dx}=\dfrac{-{{e}^{y}}}{(x{{e}^{y}}-1)}
Therefore, for the equation xey−y=5x{{e}^{y}}-y=5 , its implicit differentiation is equal to dydx=−ey(xey−1)\dfrac{dy}{dx}=\dfrac{-{{e}^{y}}}{(x{{e}^{y}}-1)}.

Note: While solving this problem, do mention the formula you are using. Avoid mathematical errors and confusion by doing step-by-step calculations. Always first solve the product rule and then solve the further equation with respect to x, to avoid mathematical errors.