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Question: Using implicit function differentiation, find \(\left( {\dfrac{{dy}}{{dx}}} \right)\)of \(\sin x + \...

Using implicit function differentiation, find (dydx)\left( {\dfrac{{dy}}{{dx}}} \right)of sin⁡x+cos⁡y=0\sin x + \cos y = 0.

Explanation

Solution

In the given problem, we are required to differentiate sin⁡x+cos⁡y=0\sin x + \cos y = 0 with respect to x. Since, sin⁡x+cos⁡y=0\sin x + \cos y = 0 is an implicit function, we will have to differentiate the function sin⁡x+cos⁡y=0\sin x + \cos y = 0 with the implicit method of differentiation. So, differentiation of sin⁡x+cos⁡y=0\sin x + \cos y = 0 with respect to x will be done layer by layer using the chain rule of differentiation as in the given function, we cannot isolate the variables x and y.

Complete step by step answer:
Consider,
sin⁡x+cos⁡y=0\sin x + \cos y = 0.
Differentiating both sides of the equation with respect to x, we get,
ddx(sin⁡x)+ddx(cos⁡y)=ddx(0)\dfrac{d}{{dx}}(\sin x) + \dfrac{d}{{dx}}\left( {\cos y} \right) = \dfrac{d}{{dx}}\left( 0 \right)
We know that derivative of sin⁡(x)\sin \left( x \right)with respect to x is cos⁡(x)\cos \left( x \right) and derivative of cos⁡(y)\cos \left( y \right) with respect to y is (−sin⁡(y))\left( { - \sin \left( y \right)} \right).
Hence, we have to apply the chain rule of differentiation in order to differentiate cos⁡(y)\cos \left( y \right) with respect to x,
=cos⁡x+(−sin⁡y).dydx=0= \cos x + ( - \sin y).\dfrac{{dy}}{{dx}} = 0
Taking cos⁡(x)\cos \left( x \right) to the right hand side so as to isolate the dydx\dfrac{{dy}}{{dx}} term, we get,
=−sin⁡ydydx=−cos⁡x= - \sin y\dfrac{{dy}}{{dx}} = - \cos x
Taking (−sin⁡(y))\left( { - \sin \left( y \right)} \right) to the right side of the equation, we get,
=dydx=−cos⁡x−sin⁡y= \dfrac{{dy}}{{dx}} = \dfrac{{ - \cos x}}{{ - \sin y}}
Cancelling the negative signs in numerator and denominator, we get,
=dydx=cos⁡xsin⁡y= \dfrac{{dy}}{{dx}} = \dfrac{{\cos x}}{{\sin y}}
So, the derivative of sin⁡x+cos⁡y=0\sin x + \cos y = 0 is dydx=cos⁡xsin⁡y\dfrac{{dy}}{{dx}} = \dfrac{{\cos x}}{{\sin y}}.
Note: Implicit functions are those functions that involve two variables and the two variables are not separable and cannot be isolated from each other. Hence, we have to follow a certain method to differentiate such functions and also apply the chain rule of differentiation. We must remember the simple derivatives of basic functions to solve such problems.