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.