Question

14. Find \(\frac{{dy}}{{dx}}\) by implicit differentiation.

14. \(\tan \left( {x - y} \right) = 2x{y^3} + 1\)

Short answer

The value is\(\frac{{dy}}{{dx}} = \frac{{{{\sec }^2}\left( {x - y} \right) - 2{y^3}}}{{{{\sec }^2}\left( {x - y} \right) + 6x{y^2}}}\).

Step-by-step solution

Differentiate the given equation

The given equation is\(\tan \left( {x - y} \right) = 2x{y^3} + 1\). Both sides of this equation are differentiated with respect to\(x\)as follows:

\(\begin{aligned}\frac{d}{{dx}}\tan \left( {x - y} \right) &= \frac{d}{{dx}}\left( {2x{y^3} + 1} \right)\\{\sec ^2}\left( {x - y} \right) \cdot \left( {1 - y'} \right) &= 2x\left( {3{y^2}y'} \right) + {y^3} \cdot 2\end{aligned}\)

Simplify the resulting equation

Simplify the resulting equation to obtain the expression for\(\frac{{dy}}{{dx}}\), as shown below:

\(\begin{aligned}{\sec ^2}\left( {x - y} \right) - y'{\sec ^2}\left( {x - y} \right) &= 6x{y^2}y' + 2{y^3}\\6x{y^2}y' + y'{\sec ^2}\left( {x - y} \right) &= {\sec ^2}\left( {x - y} \right) - 2{y^3}\\y'\left( {6x{y^2} + {{\sec }^2}\left( {x - y} \right)} \right) &= {\sec ^2}\left( {x - y} \right) - 2{y^3}\\y' &= \frac{{{{\sec }^2}\left( {x - y} \right) - 2{y^3}}}{{{{\sec }^2}\left( {x - y} \right) + 6x{y^2}}}\end{aligned}\).

Thus, the value is \(\frac{{dy}}{{dx}} = \frac{{{{\sec }^2}\left( {x - y} \right) - 2{y^3}}}{{{{\sec }^2}\left( {x - y} \right) + 6x{y^2}}}\).