# GATE Papers >> Civil >> 2019 >> Question No 40

Question No. 40

Consider two functions: $x=\psi\ln\phi$ and $y=\phi\ln\psi$. Which one of the following is the correct expression for $\frac{\partial\psi}{\partial x}$ ?

##### Answer : (D) $\frac{\ln\psi}{\ln\phi\mathrm{ln}\psi-1}$

Solution of Question No 40 of GATE 2019 Civil Paper

$\begin{array}{l}\mathrm{Given},\;\mathrm x=\mathrm\psi\mathcal l\mathrm n\mathrm\phi\;\;\&\;\;\;\mathrm y=\mathrm\phi\mathcal l\mathrm{nψ}\\\Rightarrow\mathrm\psi=\frac{\mathrm x}{\mathcal l\mathrm n\left[{\displaystyle\frac{\mathrm y}{\mathcal l\mathrm{nψ}}}\right]}\;(\begin{array}{c}\because\mathrm y=\mathrm\phi\mathcal l\mathrm{nψ}\\\Rightarrow\mathrm\phi=\frac{\mathrm y}{\mathcal l\mathrm{nψ}}\end{array}\rbrack\\\Rightarrow\mathrm\psi=\frac{\mathrm x}{\mathcal l\mathrm n\left[\mathrm y\right]-\mathcal l\mathrm n\left[\mathcal l\mathrm{nψ}\right]}\;(\because\mathcal l\mathrm n\left[\frac{\mathrm A}{\mathrm B}\right]=\mathcal l\mathrm n\left(\mathrm A\right)-\mathcal l\mathrm n\left(\mathrm B\right)\rbrack\\\therefore\mathrm{Differentiating}\;\mathrm\psi\;\mathrm{partially}\;\mathrm w.\mathrm r.\mathrm t\;'\mathrm x',\;\mathrm{we}\;\mathrm{have}\\\frac{\partial\mathrm\psi}{\partial\mathrm x}=\frac{\left[(\mathcal l\mathrm n\left(\mathrm y\right)-\mathcal l\mathrm n\left[\mathcal l\mathrm n\;\mathrm\psi\right])\left(1\right)-\mathrm x\left[0-{\displaystyle\frac1{\mathcal l\mathrm{nψ}}}{\displaystyle\frac1{\mathrm\psi}}{\displaystyle\frac{\partial\mathrm\psi}{\partial\mathrm x}}\right]\right]}{\left[\mathcal l\mathrm n\left(\mathrm y\right)-\mathcal l\mathrm n\left(\mathcal l\mathrm{nψ}\right)\right]^2};\;\mathrm{since}\;\mathrm y\;\mathrm{is}\;\mathrm{constant}\\=\frac{{\displaystyle\frac{\mathrm x}{\mathrm\psi}}+\mathrm x\left[{\displaystyle\frac1{\mathcal l\mathrm{nψ}}\frac1{\mathrm\psi}\frac{\partial\mathrm\psi}{\partial\mathrm x}}\right]}{\left({\displaystyle\frac{\mathrm x}{\mathrm\psi}}\right)2};\;\mathrm{since}\;\mathcal l\mathrm{ny}-\mathcal l\mathrm n\left(\mathcal l\mathrm{nψ}\right)=\mathcal l\mathrm n\mathrm\phi=\left[\frac{\mathrm x}{\mathrm\psi}\right]\\\Rightarrow\frac{\partial\mathrm\psi}{\partial\mathrm x}=\frac{1+{\displaystyle\frac1{\mathcal l\mathrm{nψ}}\frac{\partial\mathrm\psi}{\partial\mathrm x}}}{\displaystyle\frac{\mathrm x}{\mathrm\psi}}\Rightarrow\frac{\mathrm x}{\mathrm\psi}\frac{\partial\mathrm\psi}{\partial\mathrm x}=1+\frac1{\mathcal l\mathrm{nψ}}\frac{\partial\mathrm\psi}{\partial\mathrm x}\\\Rightarrow\frac{\mathrm x}{\mathrm\psi}\frac{\partial\mathrm\psi}{\partial\mathrm x}-\frac1{\mathcal l\mathrm{nψ}}\frac{\partial\mathrm\psi}{\partial\mathrm x}=1\\\Rightarrow\frac{\partial\mathrm\psi}{\partial\mathrm x}\left[\frac{\mathrm x}{\mathrm\psi}-\frac1{\mathcal l\mathrm{nψ}}\right]=1\Rightarrow\frac{\partial\mathrm\psi}{\partial\mathrm x}\left[\frac{\mathrm x\mathcal l\mathrm{nψ}-\mathrm\psi}{\mathrm\psi\mathcal l\mathrm{nψ}}\right]=1\\\Rightarrow\frac{\partial\mathrm\psi}{\partial\mathrm x}=\frac{\mathrm\psi\mathcal l\mathrm{nψ}}{\mathrm x\mathcal l\mathrm{nψ}-\mathrm\psi}=\frac{\mathcal l\mathrm{nψ}}{{\displaystyle\frac{\mathrm x}{\mathrm\psi}}\mathcal l\mathrm{nψ}-1}=\frac{\mathcal l\mathrm{nψ}}{\mathcal l\mathrm{n?}\mathcal l\mathrm{nψ}-1};\\\mathrm{Since}\;\frac{\mathrm x}{\mathrm\psi}=\mathcal l\mathrm n\mathrm\phi\Rightarrow\frac{\partial\mathrm\psi}{\partial\mathrm x}=\frac{\mathcal l\mathrm{nψ}}{\mathcal l\mathrm n\mathrm\phi\mathcal l\mathrm{nψ}-1}\end{array}$