5.0 Explicit and Implicit form:

When, you can write $y = f(x)$ then it is called explicit form, whereas $y + f(x)=0$ is an implicit form.

For example: Implicit equation is, $$2x^2+y^2=18$$

Then it's explicit equation will be, $$y = \pm \sqrt {18 - 2{x^2}} $$

$$y = \sqrt {18 - 2{x^2}} \,\quad and\quad y = - \sqrt {18 - 2{x^2}} $$

Question 8: Differentiate the implicit function $xy - cos(xy)= x^3y^3$

Solution: $$\frac{d}{{dx}}\left( {xy - \cos (xy)} \right) = \frac{d}{{dx}}({x^3}{y^3})$$

Using difference rule,

$$\frac{d}{{dx}}xy - \frac{d}{{dx}}\cos (xy) = \frac{d}{{dx}}({x^3}{y^3})$$

We know that, $$\frac{d}{{dx}}{x^n} = n{x^{n - 1}}$$$$\frac{d}{dx} cos x = -sin x$$

Using product rule and chain rule,

$$ y{x^{1 - 1}} + x\frac{{dy}}{{dx}} - ( - \sin (xy))\frac{d}{{dx}}(xy) = {y^3}\frac{d}{{dx}}{x^3} + {x^3}\frac{d}{{dx}}{y^3} $$$$ y + x\frac{{dy}}{{dx}} + \sin (xy)\left( {y{x^{1 - 1}} + x\frac{{dy}}{{dx}}} \right) = {y^3}3{x^{3 - 1}} + {x^3}3{y^{3 - 1}}\frac{{dy}}{{dx}} $$$$ y + x\frac{{dy}}{{dx}} + \sin (xy)\left( {y + x\frac{{dy}}{{dx}}} \right) = 3{y^3}{x^2} + 3{x^3}{y^2}\frac{{dy}}{{dx}} $$$$ y + x\frac{{dy}}{{dx}} + y\sin (xy) + x\sin (xy)\frac{{dy}}{{dx}} = 3{y^3}{x^2} + 3{x^3}{y^2}\frac{{dy}}{{dx}} $$$$ \left( {x + x\sin (xy) - 3{x^3}{y^2}} \right)\frac{{dy}}{{dx}} = 3{y^3}{x^2} - y\sin (xy) - y $$$$ \frac{{dy}}{{dx}} = \frac{{3{y^3}{x^2} - y\sin (xy) - y}}{{x + x\sin (xy) - 3{x^3}{y^2}}} $$$$ \frac{{dy}}{{dx}} = \frac{{y\left( {3{y^2}{x^2} - \sin (xy) - 1} \right)}}{{x\left( {1 + \sin (xy) - 3{x^2}{y^2}} \right)}} = - \frac{y}{x}\frac{{\left( {1 + \sin (xy) - 3{x^2}{y^2}} \right)}}{{\left( {1 + \sin (xy) - 3{x^2}{y^2}} \right)}} $$$$ \frac{{dy}}{{dx}} = - \frac{y}{x} $$

Question 9: Differentiate the implicit function $(x-y)^2= cot(2x+3y)$

Solution: $$\frac{d}{{dx}}{(x - y)^2} = \frac{d}{{dx}}(\cot (2x + 3y))$$

We know that, $$\frac{d}{{dx}}{x^n} = n{x^{n - 1}}$$$$\frac{d}{dx} cot x = -cosec ^2 x$$$$2(x - y)\frac{d}{{dx}}(x - y) = - \cos e{c^2}(2x + 3y)\frac{d}{{dx}}(2x + 3y)$$

Using sum and difference and chain rule,

$$ 2(x - y)\left( {{x^{1 - 1}} - \frac{{dy}}{{dx}}} \right) = - \cos e{c^2}(2x + 3y)\left( {2{x^{1 - 1}} + 3\frac{{dy}}{{dx}}} \right) $$$$ 2(x - y)\left( {1 - \frac{{dy}}{{dx}}} \right) = - \cos e{c^2}(2x + 3y)\left( {2 + 3\frac{{dy}}{{dx}}} \right) $$$$ 2(x - y) - 2(x - y)\frac{{dy}}{{dx}} = - 2\cos e{c^2}(2x + 3y) - 3\cos e{c^2}(2x + 3y)\frac{{dy}}{{dx}} $$$$ \left( {3\cos e{c^2}(2x + 3y) - 2(x - y)} \right)\frac{{dy}}{{dx}} = - 2\cos e{c^2}(2x + 3y) - 2(x - y) $$$$ \frac{{dy}}{{dx}} = - \frac{{2\cos e{c^2}(2x + 3y) + 2(x - y)}}{{3\cos e{c^2}(2x + 3y) - 2(x - y)}} $$

Question 10: Differentiate the implicit function $e^{xy}=e^{4x}+e^{5y}$

Solution: $$\frac{d}{{dx}}{e^{xy}} = \frac{d}{{dx}}({e^{4x}} + {e^{5y}})$$

We know that, $$\frac{d}{{dx}}{x^n} = n{x^{n - 1}}$$$$\frac{d}{dx} e^x = e^x$$

Using sum product chain and scalar rule,

$$ {e^{xy}}\frac{d}{{dx}}(xy) = {e^{4x}}\frac{d}{{dx}}(4x) + {e^{5y}}\frac{d}{{dx}}(5y) $$$$ {e^{xy}}\left( y{{x^{1 - 1}} + x\frac{{dy}}{{dx}}} \right) = 4{e^{4x}}({x^{1 - 1}}) + 5{e^{5y}}\frac{{dy}}{{dx}} $$$$ {e^{xy}}\left( {y + x\frac{{dy}}{{dx}}} \right) = 4{e^{4x}} + 5{e^{5y}}\frac{{dy}}{{dx}} $$$$ {ye^{xy}} + x{e^{xy}}\frac{{dy}}{{dx}} = 4{e^{4x}} + 5{e^{5y}}\frac{{dy}}{{dx}} $$$$ (x{e^{xy}} - 5{e^{5y}})\frac{{dy}}{{dx}} = 4{e^{4x}} - {ye^{xy}} $$$$ \frac{{dy}}{{dx}} = \frac{{4{e^{4x}} - y{e^{xy}}}}{{x{e^{xy}} - 5{e^{5y}}}} $$

Question 11: Differentiate the implicit function $3^{x}y= x\,log\,y$

Solution: $$\frac{d}{{dx}}({3^x}y) = \frac{d}{{dx}}(x{\kern 1pt} \log y)$$

We know that, $$\frac{d}{{dx}}{x^n} = n{x^{n - 1}}$$$$\frac{d}{dx} a^x = a^x log a$$$$\frac{d}{dx}{\log _e}x = \frac{1}{x ln e}= \frac{1}{x}$$

Using product rule,

$$ y\frac{d}{{dx}}{3^x} + {3^x}\frac{{dy}}{{dx}} = {x^{1 - 1}}\log y + x\frac{d}{{dx}}\log y $$$$ y{3^x}\log 3 + {3^x}\frac{{dy}}{{dx}} = \log y + \frac{x}{y}\frac{{dy}}{{dx}} $$$$ \left( {{3^x} - \frac{x}{y}} \right)\frac{{dy}}{{dx}} = \log y - {3^x}y\log 3 $$$$ \left( {\frac{{{3^x}y - x}}{y}} \right)\frac{{dy}}{{dx}} = \log y - {3^x}y\log 3 $$$$ \left( {{3^x}y - x} \right)\frac{{dy}}{{dx}} = y\left( {\log y - {3^x}y\log 3} \right) $$$$ \frac{{dy}}{{dx}} = \frac{{y\left( {\log y - {3^x}y\log 3} \right)}}{{\left( {{3^x}y - x} \right)}} $$