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Question: How do you implicitly differentiate \[ - 1 = - {y^2}x - 2xy - xy{e^x}\] ?...

How do you implicitly differentiate −1=−y2x−2xy−xyex - 1 = - {y^2}x - 2xy - xy{e^x} ?

Explanation

Solution

The above equation is an implicit equation. We can implicitly differentiate this equation by differentiating the equation with respect to xx while keeping the other variable, yy , as a function of xx . In this method, we will follow the chain rule for differentiating the function yy as we have assumed that yy is a function of xx .

Formula used:
For differentiating the above implicit equation, we follow the chain rule.
ff and gg are both differentiable functions and F(x)F\left( x \right) is the composite function which is defined by F(x)=f(g(x))F\left( x \right) = f\left( {g\left( x \right)} \right) . FF is differentiable and F′F' is the product F′(x)=f′(g(x))g′(x)F'\left( x \right) = f'\left( {g\left( x \right)} \right)g'\left( x \right) .
In the above formula, f′(g(x))f'\left( {g\left( x \right)} \right) is the outer differentiate function and g′(x)g'\left( x \right) is the inner differentiate function.

Complete step by step solution:
The equation given to us in the above question is −1=−y2x−2xy−xyex - 1 = - {y^2}x - 2xy - xy{e^x}
Now we will integrate both the sides of the equation with respect to xx . We get,
0=[−y2d(x)dx−x(d(y2))dx]−[2xd(y)dx+2yd(x)dx]−[xyd(ex)dx+xexd(y)dx+exyd(x)dx]0 = \left[ { - {y^2}\dfrac{{d\left( x \right)}}{{dx}} - x\dfrac{{\left( {d\left( {{y^2}} \right)} \right)}}{{dx}}} \right] - \left[ {2x\dfrac{{d\left( y \right)}}{{dx}} + 2y\dfrac{{d\left( x \right)}}{{dx}}} \right] - \left[ {xy\dfrac{{d\left( {{e^x}} \right)}}{{dx}} + x{e^x}\dfrac{{d\left( y \right)}}{{dx}} + {e^x}y\dfrac{{d\left( x \right)}}{{dx}}} \right]
⇒0=[−y2(1)−x(2y)dydx]−[2xdydx(1)+2y]−[xyex+xexdydx(1)+yex(1)]\Rightarrow 0 = \left[ { - {y^2}\left( 1 \right) - x\left( {2y} \right)\dfrac{{dy}}{{dx}}} \right] - \left[ {2x\dfrac{{dy}}{{dx}}\left( 1 \right) + 2y} \right] - \left[ {xy{e^x} + x{e^x}\dfrac{{dy}}{{dx}}\left( 1 \right) + y{e^x}\left( 1 \right)} \right]
On further evaluating this equation, we get,
0=−y2−2xydydx−2xdydx−2y−xyex−xexdydx−yex0 = - {y^2} - 2xy\dfrac{{dy}}{{dx}} - 2x\dfrac{{dy}}{{dx}} - 2y - xy{e^x} - x{e^x}\dfrac{{dy}}{{dx}} - y{e^x}
Now we will collect all the terms containing dydx\dfrac{{dy}}{{dx}} on one side of the equation.
−2xydydx−2xdydx−xexdydx=y2+2y+xyex+yex- 2xy\dfrac{{dy}}{{dx}} - 2x\dfrac{{dy}}{{dx}} - x{e^x}\dfrac{{dy}}{{dx}} = {y^2} + 2y + xy{e^x} + y{e^x}

⇒dydx[−2xy−2x−xex]=y2+2y+xyex+yex ⇒dydx=y2+2y+xyex+yex−2xy−2x−xex  \Rightarrow \dfrac{{dy}}{{dx}}\left[ { - 2xy - 2x - x{e^x}} \right] = {y^2} + 2y + xy{e^x} + y{e^x} \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{y^2} + 2y + xy{e^x} + y{e^x}}}{{ - 2xy - 2x - x{e^x}}} \\\

Hence after implicit differentiation, the final answer for the above equation is dydx=y2+2y+xyex+yex−2xy−2x−xex\dfrac{{dy}}{{dx}} = \dfrac{{{y^2} + 2y + xy{e^x} + y{e^x}}}{{ - 2xy - 2x - x{e^x}}} .

Note: In the above question, we have been asked to solve the equation through implicit differentiation. We have to keep this in mind that this type of differentiation can be used only when the equation consists of two independent functions and we want to derive one with respect to another. For example, in this equation we were able to derivate yy with respect to xx . Chain rule will be used to derive the function yy.