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Question: Find \(\dfrac{dy}{dx}\) if: 1\. \(\sqrt{x}+\sqrt{y}=0\) 2\. \({{e}^{x+y}}=\cos \left( x-y \righ...

Find dydx\dfrac{dy}{dx} if:
1. x+y=0\sqrt{x}+\sqrt{y}=0
2. ex+y=cos⁡(x−y){{e}^{x+y}}=\cos \left( x-y \right)
3. xy.yx=(x+y)x+y{{x}^{y}}.{{y}^{x}}={{\left( x+y \right)}^{x+y}}
4. ex−y=log⁡(xy){{e}^{x}}-y=\log \left( \dfrac{x}{y} \right)

Explanation

Solution

For solving this question you should know about the differentiation of functions. In these problems we will differentiate the given functions with respect to xx and thus we will get the solution of all these.

Complete step by step answer:
According to the problem we have to find the dydx\dfrac{dy}{dx} of all the given functions.
1. x+y=0\sqrt{x}+\sqrt{y}=0
If we differentiate this, then,
12x+12y.dydx=0 dydx.12y=−12x dydx=−12x.2y=−yx \begin{aligned} & \dfrac{1}{2\sqrt{x}}+\dfrac{1}{2\sqrt{y}}.\dfrac{dy}{dx}=0 \\\ & \dfrac{dy}{dx}.\dfrac{1}{2\sqrt{y}}=-\dfrac{1}{2\sqrt{x}} \\\ & \dfrac{dy}{dx}=-\dfrac{1}{2\sqrt{x}}.2\sqrt{y}=-\dfrac{\sqrt{y}}{\sqrt{x}} \\\ \end{aligned}
2. ex+y=cos⁡(x−y){{e}^{x+y}}=\cos \left( x-y \right)
If we differentiate it for both sides, then we get as follows,
ex+y.(1+dydx)=−sin⁡(x−y).[1−dydx] ex+y+ex+y.dydx=−sin⁡(x−y)+dydx.sin⁡(x−y) dydx(ex+y−sin⁡(x−y))=−sin⁡(x−y)−ex+y dydx=−(sin⁡(x−y)+ex+y)(−sin⁡(x−y)+ex+y) \begin{aligned} & {{e}^{x+y}}.\left( 1+\dfrac{dy}{dx} \right)=-\sin \left( x-y \right).\left[ 1-\dfrac{dy}{dx} \right] \\\ & {{e}^{x+y}}+{{e}^{x+y}}.\dfrac{dy}{dx}=-\sin \left( x-y \right)+\dfrac{dy}{dx}.\sin \left( x-y \right) \\\ & \dfrac{dy}{dx}\left( {{e}^{x+y}}-\sin \left( x-y \right) \right)=-\sin \left( x-y \right)-{{e}^{x+y}} \\\ & \dfrac{dy}{dx}=\dfrac{-\left( \sin \left( x-y \right)+{{e}^{x+y}} \right)}{\left( -\sin \left( x-y \right)+{{e}^{x+y}} \right)} \\\ \end{aligned}
3. xy.yx=(x+y)x+y{{x}^{y}}.{{y}^{x}}={{\left( x+y \right)}^{x+y}}
If we differentiate it for both sides, then we get as follows,
yx.y.xy−1+xy.xyx−1.dydx=[(x+y)(x+y)(x+y−1)]×(1+dydx) ⇒y(x+1).x(y−1)+x(y+1).y(x−1).dydx=(x+y)(x+y)+(x+y)(x+y).dydx ⇒dydx=(x+y)(x+y)−y(x+1).x(y−1)x(y+1).y(x−1)−(x+y)(x+y) \begin{aligned} & {{y}^{x}}.y.{{x}^{y-1}}+{{x}^{y}}.x{{y}^{x-1}}.\dfrac{dy}{dx}=\left[ \left( x+y \right){{\left( x+y \right)}^{\left( x+y-1 \right)}} \right]\times \left( 1+\dfrac{dy}{dx} \right) \\\ & \Rightarrow {{y}^{\left( x+1 \right)}}.{{x}^{\left( y-1 \right)}}+{{x}^{\left( y+1 \right)}}.{{y}^{\left( x-1 \right)}}.\dfrac{dy}{dx}={{\left( x+y \right)}^{\left( x+y \right)}}+{{\left( x+y \right)}^{\left( x+y \right)}}.\dfrac{dy}{dx} \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{{{\left( x+y \right)}^{\left( x+y \right)}}-{{y}^{\left( x+1 \right)}}.{{x}^{\left( y-1 \right)}}}{{{x}^{\left( y+1 \right)}}.{{y}^{\left( x-1 \right)}}-{{\left( x+y \right)}^{\left( x+y \right)}}} \\\ \end{aligned}
4. ex−y=log⁡(xy){{e}^{x}}-y=\log \left( \dfrac{x}{y} \right)
If we differentiate it for both sides, then we get as follows,
⇒ex−dydx=(yx)(−1y2.dydx) ⇒ex−dydx=−1xy.dydx ⇒(−1+1xy)dydx=−(ex) ⇒dydx(−xy+1xy)=−ex ⇒dydx=(−ex).xy1−xy \begin{aligned} & \Rightarrow {{e}^{x}}-\dfrac{dy}{dx}=\left( \dfrac{y}{x} \right)\left( \dfrac{-1}{{{y}^{2}}}.\dfrac{dy}{dx} \right) \\\ & \Rightarrow {{e}^{x}}-\dfrac{dy}{dx}=\dfrac{-1}{xy}.\dfrac{dy}{dx} \\\ & \Rightarrow \left( -1+\dfrac{1}{xy} \right)\dfrac{dy}{dx}=-\left( {{e}^{x}} \right) \\\ & \Rightarrow \dfrac{dy}{dx}\left( \dfrac{-xy+1}{xy} \right)=-{{e}^{x}} \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{\left( -{{e}^{x}} \right).xy}{1-xy} \\\ \end{aligned}

Note: While solving these types of questions you have to be careful about the power of the variable and also for the variable which is going to differentiate and thus we will get the right answers.