Question

How do you implicitly differentiate $\sin x + \cos y = \sin x\cos y$ ?

Answer 1

Hint : Differentiation calculates the rate of change of a given quantity. Implicit function definition conveys when we are not able to isolate the dependent variable in a differential equation becomes an implicit function. Both the dependent variable and independent variables are present in this type of function.

Complete step-by-step answer :
Given, $\sin x + \cos y = \sin x\cos y$ .
Doing implicit differentiation, applying differentiation on both side,
$\Rightarrow \dfrac{d}{{dx}}\left( {\sin x + \cos y} \right) = \dfrac{d}{{dx}}\left( {\sin x\cos y} \right)$
Applying sum or difference rule on the left hand side of the differential equation and Product rule on the right hand side of the differential equation we have,
$\Rightarrow \dfrac{d}{{dx}}\left( {\sin x} \right) + \dfrac{d}{{dx}}\left( {\cos y} \right) = \dfrac{d}{{dx}}\left( {\sin x\cos y} \right)$
Product rule we have $f'(x) = u'(x).v(x) + u(x).v'(x)$ . Where $u(x) = \sin x$ and $v(x) = \cos y$ .
$\Rightarrow \dfrac{d}{{dx}}\left( {\sin x} \right) + \dfrac{d}{{dx}}\left( {\cos y} \right) = \dfrac{d}{{dx}}(\sin x)\cos y + \sin x\dfrac{d}{{dx}}(\cos y)$
We know that, $\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x$ and $\dfrac{d}{{dx}}(\cos y) = - \sin y\dfrac{{dy}}{{dx}}$ . Substituting this we have,
$\Rightarrow \cos x + \left( { - \sin y\dfrac{{dy}}{{dx}}} \right) = \cos x\cos y + \sin x\left( { - \sin y\dfrac{{dy}}{{dx}}} \right)$
$\Rightarrow \cos x - \sin y\dfrac{{dy}}{{dx}} = \cos x\cos y - \sin y\sin x\dfrac{{dy}}{{dx}}$ .
Regrouping $\dfrac{{dy}}{{dx}}$ terms we get,
$\Rightarrow \sin y\sin x\dfrac{{dy}}{{dx}} - \sin y\dfrac{{dy}}{{dx}} = \cos x\cos y - \cos x$
Taking $\dfrac{{dy}}{{dx}}$ common on the left hand side of the equation,
$\Rightarrow \dfrac{{dy}}{{dx}}\left( {\sin y\sin x - \sin y} \right) = \cos x\cos y - \cos x$
Now dividing $\left( {\sin y\sin x - \sin y} \right)$ on both sides of the problem we have,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\cos x\cos y - \cos x}}{{\sin y\sin x - \sin y}}$
Taking $\cos x$ common in the numerator and taking $\sin y$ common in the denominator term we have,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\cos x(\cos y - 1)}}{{\sin y(\sin x - 1)}}$ . This is the required answer.
So, the correct answer is “ $\dfrac{{dy}}{{dx}} = \dfrac{{\cos x(\cos y - 1)}}{{\sin y(\sin x - 1)}}$ ”.

Note : We have different types of differentiation rules.
Sum or difference rule: when the function is the sum or difference of two functions, the derivative is the sum or difference of derivative of each function. i.e.,
$f(x) = u(x) \pm v(x)$ then $f'(x) = u'(x) \pm v'(x)$ .
Product rule: when f(x) is the product of two function that is $f(x) = u(x).v(x)$ then $f'(x) = u'(x).v(x) + u(x).v'(x)$ .
Quotient rule: , if the two functions $f(x)$ and $g(x)$ are differentiable then the quotient is differentiable and ${\left( {\dfrac{{f(x)}}{{g(x)}}} \right)^1} = \dfrac{{f'(x)g(x) - f(x)g'(x)}}{{{{\left( {g(x)} \right)}^2}}}$ .
We use this rule depending on the given problem. Careful in the differentiation a function f(y) with respect to x. we will have $\dfrac{d}{{dx}}(f(y)) = f'(y)\dfrac{{dy}}{{dx}}$ .