Question

If $\tan \left( {x + y} \right) + \tan \left( {x - y} \right) = 1$ , then find $\dfrac{{dy}}{{dx}}$ .

Answer 1

Hint: It is asked to find $\dfrac{{dy}}{{dx}}$ of the equation $\tan \left( {x + y} \right) + \tan \left( {x - y} \right) = 1$ .
So, differentiate the whole equation with respect to x.
Now, on applying chain rule for differentiation in L.H.S. of the equation, we will get $\dfrac{{dy}}{{dx}}$ .

Complete step-by-step answer:
The given equation is $\tan \left( {x + y} \right) + \tan \left( {x - y} \right) = 1$ .
Now, we are asked to find $\dfrac{{dy}}{{dx}}$ .
For that, we will differentiate the given equation with respect to x.
Now, differentiating with respect to x,
 $
  \therefore \dfrac{d}{{dx}}\left( {\tan \left( {x + y} \right) + \tan \left( {x - y} \right)} \right) = \dfrac{d}{{dx}}\left( 1 \right) \\
  \therefore \dfrac{d}{{dx}}\tan \left( {x + y} \right) + \dfrac{d}{{dx}}\tan \left( {x - y} \right) = 0 \\
 $
Here, we will have to use chain rules for differentiating.
 $
  \Rightarrow {\sec ^2}\left( {x + y} \right)\left[ {1 + \dfrac{{dy}}{{dx}}} \right] + {\sec ^2}\left( {x - y} \right)\left[ {1 - \dfrac{{dy}}{{dx}}} \right] = 0 \\
  \Rightarrow {\sec ^2}\left( {x + y} \right) + {\sec ^2}\left( {x + y} \right)\dfrac{{dy}}{{dx}} + {\sec ^2}\left( {x - y} \right) - {\sec ^2}\left( {x - y} \right)\dfrac{{dy}}{{dx}} = 0 \\
  \Rightarrow {\sec ^2}\left( {x + y} \right)\dfrac{{dy}}{{dx}} - {\sec ^2}\left( {x - y} \right)\dfrac{{dy}}{{dx}} = - \left[ {{{\sec }^2}\left( {x + y} \right) + {{\sec }^2}\left( {x - y} \right)} \right] \\
  \Rightarrow \left[ {{{\sec }^2}\left( {x + y} \right) - {{\sec }^2}\left( {x - y} \right)} \right]\dfrac{{dy}}{{dx}} = - \left[ {{{\sec }^2}\left( {x + y} \right) + {{\sec }^2}\left( {x - y} \right)} \right] \\
  \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - \left[ {{{\sec }^2}\left( {x + y} \right) + {{\sec }^2}\left( {x - y} \right)} \right]}}{{\left[ {{{\sec }^2}\left( {x + y} \right) - {{\sec }^2}\left( {x - y} \right)} \right]}} \\
  \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\left[ {{{\sec }^2}\left( {x + y} \right) + {{\sec }^2}\left( {x - y} \right)} \right]}}{{\left[ {{{\sec }^2}\left( {x - y} \right) - {{\sec }^2}\left( {x + y} \right)} \right]}} \\
 $
Thus, we get $\dfrac{{dy}}{{dx}} = \dfrac{{\left[ {{{\sec }^2}\left( {x + y} \right) + {{\sec }^2}\left( {x - y} \right)} \right]}}{{\left[ {{{\sec }^2}\left( {x - y} \right) - {{\sec }^2}\left( {x + y} \right)} \right]}}$ .

Note: Chain rule for differentiation:
To find the derivative of a composite function, we use chain rule. The chain rule tells us how to differentiate composite functions.
Let us consider a composite function \[f\left( {g\left( x \right)} \right)\] . Here, g is the function within the function f, so, we will call f as outer function and g as inner function.
Now, to find its derivative, first we differentiate the outer function i.e. function f and then add the derivative of the inner function i.e. function g to it to get the derivative of the composite function.
 $
  \dfrac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right] = \dfrac{d}{{dx}}f\left( {g\left( x \right)} \right) + \dfrac{d}{{dx}}g\left( x \right) \\
   = f'\left( {g\left( x \right)} \right) + g'\left( x \right) \\
 $