What's in the denominator?

Calculus Level 3

Given that for $x>0$ , $y=y(x)>0$ , $e^{-\cot(y)}=x^x+x\ln(x)$

and implicit differentiation of the above equation gives $\frac{dy}{dx}=\frac{(x^x+1)(\ln(x)+1)}{(\csc^2(y))(x^x+xf(x))}$

where $f(x)$ is a function involving $x$ (not necessarily equalling $y$ ), find $f(x)$ .

e^x

\sqrt{x^2+1}

\sin(x)

\tan^2(x)

\frac{1}{x}

\ln(x)

\sec(x)

1 solution

Wee Xian Bin
Aug 13, 2016

Implicit differentiation of $e^{-\cot(y)}=x^x+x\ln(x)$ gives $\frac{dy}{dx} (e^{-\cot(y)}) (\csc^2(y)) = x^x(\ln(x)+1) + \ln(x)+1$ $\frac{dy}{dx} (e^{-\cot(y)}) (\csc^2(y)) = (x^x+1)(\ln(x)+1)$ rearrangement gives $\frac{dy}{dx} = \frac{(x^x+1)(\ln(x)+1)}{(\csc^2(y))(e^{-\cot(y)})}$ substitution back the expression of $e^{-\cot(y)}$ gives $\frac{dy}{dx} = \frac{(x^x+1)(\ln(x)+1)}{(\csc^2(y))(x^x+x\ln(x))}$ hence $f(x)=\ln(x)$

Addendum: To differentiate $x^x$ , consider $z=z(x)=x^x$ $\ln{(z)}=x\ln{(x)}$ then differentiate both sides implicitly $\frac{(\frac{dz}{dx})}{z}=\ln(x)+1$ therefore $\frac{dz}{dx}=z(\ln{(x)}+1)= x^x(\ln{(x)}+1)$

What's in the denominator?

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