Question

How do you find the derivative of $x + \tan (xy) = 0$ ?

Answer 1

In this question we can use the Product Rule method and the Chain Rule method in differentiation. In differentiation, product rule is used for differentiating questions where one function is getting multiplied by another function. The Chain rule is used to find the derivation of a number which is composite in nature.

Complete step by step solution:
Our question is to find the derivative of $x + \tan (xy) = 0$. We will first differentiate from both the sides, and then we get:
$\dfrac{d}{{dx}}(x) + \dfrac{d}{{dx}}(\tan (xy)) = \dfrac{d}{{dx}}(0)$
Now, we know that $\dfrac{d}{{dx}}(x) = 1\,;\,\dfrac{d}{{dx}}(0) = 0$, from the basic formulas of differentiation. So, we can rewrite the equation as:
$\Rightarrow 1 + \dfrac{d}{{dx}}(\tan (xy)) = 0$

Now, we have to find the derivation of $\tan (xy)$. We know that the derivative of $\tan$ is ${\sec ^2}$. We have to apply the Chain Rule here, and then we get:
$\dfrac{d}{{dx}}(\tan (xy)) = {\sec ^2}(xy) \cdot \dfrac{{d(xy)}}{{dx}}$
Now we have to put the derivative of $\tan (xy)$ in $1 + \dfrac{d}{{dx}}(\tan (xy)) = 0$, and then we get:
$\Rightarrow 1 + {\sec ^2}(xy) \cdot \dfrac{{d(xy)}}{{dx}} = 0$
Now, we will apply the Product Rule here for $\dfrac{{d(xy)}}{{dx}}$, and we get:
$\Rightarrow \dfrac{{d(xy)}}{{dx}} = x\dfrac{{dy}}{{dx}} + y\dfrac{{dx}}{{dx}}$
Here, the similar terms get cancelled and we get:
$\Rightarrow \dfrac{{d(xy)}}{{dx}} = x\dfrac{{dy}}{{dx}} + y$

Now, we will put the value of $\dfrac{{d(xy)}}{{dx}}$ in the equation $1 + {\sec ^2}(xy) \cdot \dfrac{{d(xy)}}{{dx}} = 0$, and we get:
$\Rightarrow 1 + {\sec ^2}(xy) \cdot (x\dfrac{{dy}}{{dx}} + y) = 0$
Next, we will start simplifying the equation. First, we will subtract $1$ from both the sides:
$\Rightarrow 1 + {\sec ^2}(xy) \cdot (x\dfrac{{dy}}{{dx}} + y) - 1 = 0 - 1$
$\Rightarrow {\sec ^2}(xy) \cdot (x\dfrac{{dy}}{{dx}} + y) = - 1$
Now, we will divide ${\sec ^2}(xy)$ on both the sides and, we will get:
$\Rightarrow (x\dfrac{{dy}}{{dx}} + y) = \dfrac{{ - 1}}{{{{\sec }^2}(xy)}}$

Now, we know that $\dfrac{1}{{{{\sec }^2}}} = {\cos ^2}$. Therefore, $\dfrac{1}{{{{\sec }^2}(xy)}} = {\cos ^2}(xy)$
When we put this value of $\dfrac{1}{{{{\sec }^2}(xy)}}$ in the equation $(x\dfrac{{dy}}{{dx}} + y) = \dfrac{{ - 1}}{{{{\sec }^2}(xy)}}$, then we get:
$\Rightarrow (x\dfrac{{dy}}{{dx}} + y) = - {\cos ^2}(xy)$
When we shift $y$ to the other side, then we get:
$\Rightarrow x\dfrac{{dy}}{{dx}} = - {\cos ^2}(xy) - y$
By dividing $x$ on both the sides we get:
$\therefore\dfrac{{dy}}{{dx}} = \dfrac{{ - {{\cos }^2}(xy) - y}}{x}$

So, the final answer is that the derivative of $x + \tan (xy) = 0$ is $- \dfrac{{{{\cos }^2}(xy) - y}}{x}$.

Note: The above method is easy to solve. There is another method to get the derivative of $x + \tan (xy) = 0$. We get that by using the Quotient Rule. In that we first shift the $\tan$ part to the other side and apply the Quotient Rule. This is also a very easy process.