21.9 - Derivatives of Implicit Functions - Easy method

In the previous section, we saw the method to find the derivatives of implicit functions. In this section, we will see some solved examples. In those solved examples, we will see an easy method also.

Solved example 21.32
Find $\frac{dy}{dx}$ if x + sin (xy) − y = 0.
Solution:
1. Consider the function x + sin (xy) − y = 0.
• We want the derivative of this function with respect to x.
   ♦ That is., we want $\frac{dy}{dx}$.
2. Consider the "y" in the given equation. It is a function of "x" because, value of y depends on the value of x. So we can write "f(x)" in the place of "y". We get:
x + sin (x f(x)) − f(x) = 0
(We need not worry about how f(x) is defined. This fact will become clear when we solve a few problems)
3. Differentiating both sides, we get:
$\frac{d}{dx} (x) \,+\, \frac{d}{dx} \sin(x f(x)) \, - \, \frac{d}{dx} f(x) \,=\, \frac{d}{dx} (0)$
⇒ $1 \,+\, \frac{d}{dx} \sin(x f(x)) \, - \, f'(x) \,=\, 0$
4. The second term in the L.H.S needs special attention. It is the derivative of a composite function. It can be determined in 4 steps:
(i) g(x) = sin (x f(x)) = (v∘u)(x) = v(u(x))
Where u(x) = x f(x) and v(u(x)) = sin (x f(x)).
(ii) v'(u(x)) = cos (x f(x))
(iii) u'(x) = 1.f(x) + x f'(x)
(iv) So g'(x) = v'(u(x)).u'(x)
= f(x).cos (x f(x)) + x f'(x) cos(x f(x))
5. So the result in (3) becomes:
$1 \,+\, f(x).\cos (x f(x)) + x. f'(x). \cos(x f(x)) \, - \, f'(x) \,=\, 0$
⇒ $1 \,+\, f(x).\cos (x f(x)) ~+~  f'(x) \left[x. \cos(x. f(x)) \, - \, 1 \right] \,=\, 0$
⇒ $1 \,+\, f(x).\cos (x f(x)) ~-~  f'(x) \left[1~-~x. \cos(x. f(x))  \right] \,=\, 0$
⇒ $f'(x) = \frac{1 \,+\, f(x).\cos (x. f(x))}{1~-~x. \cos(x. f(x))}$
6. We wrote "f(x)" in the place of "y".
• So f'(x) is $\frac{dy}{dx}$
• Therefore, $\frac{dy}{dx} = \frac{1 \,+\, y \cos (xy)}{1~-~x \cos(xy)}$

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Once we become familiar with the basics of this method,
    ♦ we need not write "f(x)" in the place of "y". 
        ✰ we can keep "y" as such.
    ♦ we need not write "f'(x)" to show differentiation of y.
        ✰ we can write $\frac{dy}{dx}$ instead.
    ♦ Also in simple cases, we may be able to apply chain rule in just one step.

So let us redo the above solved example.

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{d}{dx} (x)\,+\,\frac{d}{dx} (\sin(xy))\,-\,\frac{d}{dx} (y)}    & {~=~}    &{\frac{d}{dx} (0)}    \\
{~\color{magenta}    2    }    &{\implies}    &{1\,+\,\frac{d}{dx} (\sin(xy))\,-\,\frac{dy}{dx}}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{\implies}    &{1\,+\,\cos(xy) \left[\frac{d}{dx} (x) .y + x.\frac{dy}{dx} \right]\,-\,\frac{dy}{dx}}    & {~=~}    &{0}    \\
{~\color{magenta}    4    }    &{\implies}    &{1\,+\,\cos(xy) \left[y + x.\frac{dy}{dx} \right]\,-\,\frac{dy}{dx}}    & {~=~}    &{0}    \\
{~\color{magenta}    5    }    &{\implies}    &{1\,+\,y \cos(xy) + x.\frac{dy}{dx} \cos(xy)\,-\,\frac{dy}{dx}}    & {~=~}    &{0}    \\
{~\color{magenta}    6    }    &{\implies}    &{1\,+\,y \cos(xy) + \frac{dy}{dx} \left[x \cos(xy)\,-\,1 \right]}    & {~=~}    &{0}    \\
{~\color{magenta}    7    }    &{\implies}    &{1\,+\,y \cos(xy)}    & {~=~}    &{\frac{dy}{dx} \left[1 - x \cos(xy) \right]}    \\
{~\color{magenta}    8    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{1\,+\,y \cos(xy)}{1 - x \cos(xy)}}    \\
\end{array}$

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Solved example 21.33
Find $\frac{dy}{dx}$ if x³y⁵ + 3x = 8y³ + 1.
Solution:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{d}{dx} (x^3 y^5)\,+\,\frac{d}{dx} (3x)}    & {~=~}    &{\frac{d}{dx} (8 y^3)\,+\,\frac{d}{dx} (1)}    \\
{~\color{magenta}    2    }    &{\implies}    &{\frac{d}{dx} (x^3 y^5)\,+\,3}    & {~=~}    &{\frac{d}{dx} (8 y^3)\,+\,0}    \\
{~\color{magenta}    3    }    &{\implies}    &{3 x^2 . y^5 \,+\,x^3 . \frac{d}{dx} (y^5) \,+\,3}    & {~=~}    &{8 \frac{d}{dx} (y^3)}    \\
{~\color{magenta}    4    }    &{\implies}    &{3 x^2 . y^5 \,+\,x^3 . \left[5 y^4 .  \frac{dy}{dx} \right] \,+\,3}    & {~=~}    &{8. 3 y^2 . \frac{dy}{dx}}    \\
{~\color{magenta}    5    }    &{\implies}    &{3 x^2 y^5 \,+\, 5 x^3 y^4 \frac{dy}{dx} \,+\, 3}    & {~=~}    &{24 y^2 \frac{dy}{dx}}    \\
{~\color{magenta}    6    }    &{\implies}    &{3 x^2 y^5  \,+\, 3}    & {~=~}    &{24 y^2 \frac{dy}{dx} \,-\, 5 x^3 y^4 \frac{dy}{dx}}    \\
{~\color{magenta}    7    }    &{\implies}    &{3 x^2 y^5  \,+\, 3}    & {~=~}    &{\left[24 y^2  \,-\, 5 x^3 y^4 \right]\frac{dy}{dx}}    \\
{~\color{magenta}    8    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{3 x^2 y^5  \,+\, 3}{24 y^2  \,-\, 5 x^3 y^4}}    \\
\end{array}$

◼ Remarks:
• 3 (magenta color). For differentiating x³y⁵, we use product rule. For differentiating y⁵, we treat it as a composite function.

• In the R.H.S, y³ is also treated as a composite function.

Solved example 21.34
Find $\frac{dy}{dx}$ if x² + xy + y² = 100.
Solution:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{d}{dx} (x^2)\,+\,\frac{d}{dx} (xy) \,+\,\frac{d}{dx} (y^2)}    & {~=~}    &{\frac{d}{dx} (100)}    \\
{~\color{magenta}    2    }    &{\implies}    &{2x\,+\,\frac{d}{dx} (xy) \,+\,\frac{d}{dx} (y^2)}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{\implies}    &{2x\,+\,\left[1. {y} + x \frac{dy}{dx} \right] \,+\,\left[2y \frac{dy}{dx} \right]}    & {~=~}    &{0}    \\
{~\color{magenta}    4    }    &{\implies}    &{2x + y + (x+2y)\frac{dy}{dx}}    & {~=~}    &{0}    \\
{~\color{magenta}    5    }    &{\implies}    &{2x + y}    & {~=~}    &{- (x+2y)\frac{dy}{dx}}    \\
{~\color{magenta}    6    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{-(2x + y)}{(x+2y)}}    \\
\end{array}$

Solved example 21.35
Find $\frac{dy}{dx}$ if x³ + x²y + xy² + y³ = 81.
Solution:

Solved example 21.36
Find $\frac{dy}{dx}$ if sin²y + cos xy = π.
Solution:

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We have completed a discussion on the derivatives of implicit functions. In the next section, we will see derivatives of inverse trigonometric functions.

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