Question

If
${ \tan( {x}^{2} + {y}^{2} ) }^{t} = {e}^{x}$
Find dy/dx​

Answer 1

Given,

$\longrightarrow\sf{\tan[(x^2+y^2)^t]=e^x}$

$\longrightarrow\sf{(x^2+y^2)^t=\tan^{-1}(e^x)}$

$\longrightarrow\sf{x^2+y^2=\left[\tan^{-1}(e^x)\right]^{\frac{1}{t}}}$

$\longrightarrow\sf{y^2=\left[\tan^{-1}(e^x)\right]^{\frac{1}{t}}-x^2}$

$\longrightarrow\sf{y=\left[\left[\tan^{-1}(e^x)\right]^{\frac{1}{t}}-x^2\right]^{\frac{1}{2}}}$

Hence,

$\longrightarrow\sf{\dfrac{dy}{dx}=\dfrac{d}{dx}\left[\left[\tan^{-1}(e^x)\right]^{\frac{1}{t}}-x^2\right]^{\frac{1}{2}}}$

We know the two rules,

• $\sf{\dfrac{d}{dx}\left[f(g(x))\right]=\dfrac{d\,[f(g(x))]}{d\,[g(x)]}\cdot\dfrac{d\,[g(x)]}{dx}}$

• $\sf{\dfrac{d}{dx}[x^n]=nx^{n-1}}$

Then,

$\longrightarrow\sf{\dfrac{dy}{dx}=\dfrac{1}{2}\left[\left[\tan^{-1}(e^x)\right]^{\frac{1}{t}}-x^2\right]^{\frac{1}{2}-1}\cdot\dfrac{d}{dx}\left[\left[\tan^{-1}(e^x)\right]^{\frac{1}{t}}-x^2\right]}$

$\longrightarrow\sf{\dfrac{dy}{dx}=\dfrac{1}{2\left[\left[\tan^{-1}(e^x)\right]^{\frac{1}{t}}-x^2\right]^{\frac{1}{2}}}\cdot\left[\dfrac{d}{dx}\left[\tan^{-1}(e^x)\right]^{\frac{1}{t}}-2x\right]}$

$\longrightarrow\sf{\dfrac{dy}{dx}=\dfrac{\left[\dfrac{1}{t}\left[\tan^{-1}(e^x)\right]^{\frac{1}{t}-1}\cdot\dfrac{d}{dx}\left[\tan^{-1}(e^x)\right]-2x\right]}{2\left[\left[\tan^{-1}(e^x)\right]^{\frac{1}{t}}-x^2\right]^{\frac{1}{2}}}}$

Since $\sf{\dfrac{d}{dx}[\tan^{-1}(x)]=\dfrac{1}{x^2+1},}$

$\longrightarrow\sf{\dfrac{dy}{dx}=\dfrac{\left[\dfrac{1}{t}\left[\tan^{-1}(e^x)\right]^{\frac{1-t}{t}}\cdot\dfrac{1}{(e^x)^2+1}\cdot\dfrac{d}{dx}(e^x)-2x\right]}{2\left[\left[\tan^{-1}(e^x)\right]^{\frac{1}{t}}-x^2\right]^{\frac{1}{2}}}}$

And since $\sf{\dfrac{d}{dx}(e^x)=e^x,}$

$\longrightarrow\sf{\dfrac{dy}{dx}=\dfrac{\left[\dfrac{1}{t}\left[\tan^{-1}(e^x)\right]^{\frac{1-t}{t}}\cdot\dfrac{1}{e^{2x}+1}\cdot e^x-2x\right]}{2\left[\left[\tan^{-1}(e^x)\right]^{\frac{1}{t}}-x^2\right]^{\frac{1}{2}}}}$

$\longrightarrow\sf{\dfrac{dy}{dx}=\dfrac{\dfrac{e^x\left[\tan^{-1}(e^x)\right]^{\frac{1-t}{t}}}{t\left[e^{2x}+1\right]}-2x}{2\left[\left[\tan^{-1}(e^x)\right]^{\frac{1}{t}}-x^2\right]^{\frac{1}{2}}}}$

$\longrightarrow\underline{\underline{\sf{\dfrac{dy}{dx}=\dfrac{e^x\left[\tan^{-1}(e^x)\right]^{\frac{1-t}{t}}-2tx\left[e^{2x}+1\right]}{2t\left[e^{2x}+1\right]\left[\left[\tan^{-1}(e^x)\right]^{\frac{1}{t}}-x^2\right]^{\frac{1}{2}}}}}}$