# How do you implicitly differentiate  (xy)^2+cos(xy)=ln(xy)+tanh(xy) ?

##### 1 Answer

$\textcolor{red}{y ' = \frac{\frac{1}{x} - 2 x {y}^{2} + y \cdot \sin \left(x y\right) + y \cdot {\sech}^{2} \left(x y\right)}{- \frac{1}{y} + 2 {x}^{2} y - x \cdot \sin \left(x y\right) - x \cdot {\sech}^{2} \left(x y\right)}}$

#### Explanation:

We start from the given

${\left(x y\right)}^{2} + \cos \left(x y\right) = \ln \left(x y\right) + \tanh \left(x y\right)$

differentiate both sides of the equation

$\frac{d}{\mathrm{dx}} {\left(x y\right)}^{2} + \frac{d}{\mathrm{dx}} \cos \left(x y\right) = \frac{d}{\mathrm{dx}} \ln \left(x y\right) + \frac{d}{\mathrm{dx}} \tanh \left(x y\right)$

$2 \left(x y\right) \cdot \left(x y ' + y \cdot 1\right) + \left(- \sin \left(x y\right) \cdot \left(x y ' + y \cdot 1\right)\right) = \frac{1}{x y} \cdot \left(x y ' + y \cdot 1\right) + {\sech}^{2} \left(x y\right) \cdot \left(x y ' + y \cdot 1\right)$

$2 {x}^{2} y y ' + 2 x {y}^{2} - x \cdot \sin \left(x y\right) \cdot y ' - y \cdot \sin \left(x y\right) = \frac{y '}{y} + \frac{1}{x} + x \cdot {\sech}^{2} \left(x y\right) \cdot y ' + y \cdot {\sech}^{2} \left(x y\right)$

$2 {x}^{2} y y ' - x \cdot \sin \left(x y\right) \cdot y ' - \frac{y '}{y} - x \cdot {\sech}^{2} \left(x y\right) \cdot y ' = \frac{1}{x} + y \cdot {\sech}^{2} \left(x y\right) - 2 x {y}^{2} + y \cdot \sin \left(x y\right)$

$\left(2 {x}^{2} y - x \cdot \sin \left(x y\right) - \frac{1}{y} - x \cdot {\sech}^{2} \left(x y\right)\right) y ' = \frac{1}{x} + y \cdot {\sech}^{2} \left(x y\right) - 2 x {y}^{2} + y \cdot \sin \left(x y\right)$

$\textcolor{red}{y ' = \frac{\frac{1}{x} - 2 x {y}^{2} + y \cdot \sin \left(x y\right) + y \cdot {\sech}^{2} \left(x y\right)}{- \frac{1}{y} + 2 {x}^{2} y - x \cdot \sin \left(x y\right) - x \cdot {\sech}^{2} \left(x y\right)}}$

God bless....I hope the explanation is useful.