Question

How do you find $\dfrac{dy}{dx}$ by implicit differentiation of $\tan \left( x+y \right)=x$ and evaluate at point $\left( 0,0 \right)$?

Answer 1

This question is from the topic of calculus. In this question, we are going to differentiate the given equation that $\tan \left( x+y \right)=x$ by implicit differentiation. We will first differentiate the whole equation with respect to x. After that, we will use the formula $\dfrac{d}{dx}\tan x={{\sec }^{2}}x$ and do the further differentiation. In this question, we are going to use chain rule. After differentiating further, we will make one side of the equation $\dfrac{dy}{dx}$ and then equate this with the other side of the equation.

Complete step by step solution:
Let us solve this question.
In this question, it is asked us to find $\dfrac{dy}{dx}$ by implicit differentiation of $\tan \left( x+y \right)=x$ at point (0,0).
The term we have to differentiate is
$\tan \left( x+y \right)=x$
Now, differentiating both side of the above term with respect to x, we can write
$\Rightarrow \dfrac{d}{dx}\tan \left( x+y \right)=\dfrac{d}{dx}x$
As we know that differentiation of the term x with respect to x is 1, so we can write
$\Rightarrow \dfrac{d}{dx}\tan \left( x+y \right)=1$
Now, using the formula $\dfrac{d}{dx}\tan x={{\sec }^{2}}x$, we can write
$\Rightarrow {{\sec }^{2}}\left( x+y \right)\dfrac{d}{dx}\left( x+y \right)=1$
We have used chain rule here in the above differentiation.
We can write the above differentiation as
$\Rightarrow {{\sec }^{2}}\left( x+y \right)\left( \dfrac{d}{dx}x+\dfrac{d}{dx}y \right)=1$
The above differentiation can also be written as
$\Rightarrow {{\sec }^{2}}\left( x+y \right)\left( 1+\dfrac{dy}{dx} \right)=1$
Now, taking the ‘sec’ term to the right side of the equation, we get
$\Rightarrow \left( 1+\dfrac{dy}{dx} \right)=\dfrac{1}{{{\sec }^{2}}\left( x+y \right)}$
The above equation can also be written as
$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{{{\sec }^{2}}\left( x+y \right)}-1$
We can write the above equation as
$\Rightarrow \dfrac{dy}{dx}=\dfrac{1-{{\sec }^{2}}\left( x+y \right)}{{{\sec }^{2}}\left( x+y \right)}$
Now, we have found the value of $\dfrac{dy}{dx}$ by implicit differentiation. The value of $\dfrac{dy}{dx}$ is $\dfrac{1-{{\sec }^{2}}\left( x+y \right)}{{{\sec }^{2}}\left( x+y \right)}$.

Note:
We should have a better knowledge in the topic of calculus. We should know about chain rule. The chain rule helps us to differentiate composite functions. The composite functions are always in the form of f(g(x)). The chain rule states that if we differentiate composite function f(g(x)) then $\dfrac{d}{dx}\text{f(g(x))}$ will be like we will first differentiate the function f and then we will differentiate the function g, where f and g are two functions. For example, we can see below:
$\dfrac{d}{dx}\sin {{x}^{2}}=\cos {{x}^{2}}\dfrac{d}{dx}{{x}^{2}}=\left( \cos {{x}^{2}} \right)2x=2x\cos {{x}^{2}}$
We should remember the formula:
$\dfrac{d}{dx}\tan x={{\sec }^{2}}x$