Question

How do you implicitly differentiate $- 1 = x - y\tan (x + 2y)$ ?

Answer 1

Hint : In this question we are asked to differentiate a function implicitly. In order to proceed with this question we need to know how to differentiate implicitly. The $y$ is not completely expressed in terms of $x$ then the function is called an explicit function. The method of implicit differentiation allows you to find the derivative of $y$ with respect to x without the need to solve the given equation for $y$ .
To solve this question first differentiate both the sides with respect to $x$ . Then collect all the terms having $\dfrac{{dy}}{{dx}}$ on one side and all the other terms on the other side. Then we need to factor $\dfrac{{dy}}{{dx}}$ out of the left side of the equation by dividing the coefficient of $\dfrac{{dy}}{{dx}}$ to get required $\dfrac{{dy}}{{dx}}$ . Then finally solve for $\dfrac{{dy}}{{dx}}$ .

Complete step by step solution:
We are given,
$- 1 = x - y\tan (x + 2y)$
We’ll differentiate with respect to x
$\Rightarrow 0 = 1 - [y \times {\sec ^2}(x + 2y) \times (1 + 2\dfrac{{dy}}{{dx}}) + \tan (x + 2y) \times \dfrac{{dy}}{{dx}}]$
We’ll send $1$ to the other side,
$\Rightarrow 1 = [y \times {\sec ^2}(x + 2y) \times (1 + 2\dfrac{{dy}}{{dx}}) + \tan (x + 2y) \times \dfrac{{dy}}{{dx}}]$
$\Rightarrow 1 = [y \times {\sec ^2}(x + 2y) \times (1 + 2\dfrac{{dy}}{{dx}}) + \tan (x + 2y) \times \dfrac{{dy}}{{dx}}]$
$\Rightarrow 1 = [y{\sec ^2}(x + 2y) + 2y{\sec ^2}(x + 2y)\dfrac{{dy}}{{dx}} + \tan (x + 2y) \times \dfrac{{dy}}{{dx}}]$
Now collect all the terms containing $\dfrac{{dy}}{{dx}}$ on one side,
$\Rightarrow 1 - y{\sec ^2}(x + 2y) = 2y{\sec ^2}(x + 2y)\dfrac{{dy}}{{dx}} + \tan (x + 2y) \times \dfrac{{dy}}{{dx}}$
$\Rightarrow 1 - y{\sec ^2}(x + 2y) = [2y{\sec ^2}(x + 2y) + \tan (x + 2y)]\dfrac{{dy}}{{dx}}$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{1 - y{{\sec }^2}(x + 2y)}}{{[2y{{\sec }^2}(x + 2y) + \tan (x + 2y)}}$
This is the required answer
So, the correct answer is “$\dfrac{{dy}}{{dx}} = \dfrac{{1 - y{{\sec }^2}(x + 2y)}}{{[2y{{\sec }^2}(x + 2y) + \tan (x + 2y)}}$”.

Note : When we differentiate a function with $y$ , first we differentiate y and then write $\dfrac{{dy}}{{dx}}$ every time we differentiate $y$ . When y is the function of $x$ and terms contain $y$ , then differentiate the function same as the composite function. The other type of function is explicit function, where $y$ can be completely expressed in terms of $x$ . This implicit equation defines f as a function of $x$ only if $- 1{\text{ }} \leqslant \;x\; \leqslant {\text{ }}1\;$ and one considers only non-negative (or non-positive) values for the values of the function. It is important to know chain rules to deal with functions implicitly.