Question

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

19. \(\sqrt {x + y} = {x^4} + {y^4}\)

Short answer

The value is \(\frac{{dy}}{{dx}} = \frac{{1 - 8{x^3}\sqrt {x + y} }}{{8{y^3}\sqrt {x + y} - 1}}\).

Step-by-step solution

Differentiate the given equation

The given equation is\(\sqrt {x + y} = {x^4} + {y^4}\). Both sides of this equation are differentiated with respect to\(x\)as follows:

\(\begin{aligned}\frac{d}{{dx}}\left( {\sqrt {x + y} } \right) &= \frac{d}{{dx}}\left( {{x^4} + {y^4}} \right)\\\frac{1}{2}{\left( {x + y} \right)^{ - \frac{1}{2}}}\left( {1 + y'} \right) &= 4{x^3} + 4{y^3}y'\end{aligned}\)

Simplify the resulting equation

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

\(\begin{aligned}\frac{1}{{2\sqrt {x + y} }} + \frac{1}{{2\sqrt {x + y} }}y' &= 4{x^3} + 4{y^3}y'\\\frac{1}{{2\sqrt {x + y} }} - 4{x^3} &= 4{y^3}y' - \frac{1}{{2\sqrt {x + y} }}y'\\\frac{{1 - 8{x^3}\sqrt {x + y} }}{{2\sqrt {x + y} }} &= \frac{{8{y^3}\sqrt {x + y} - 1}}{{2\sqrt {x + y} }}y'\\y' &= \frac{{1 - 8{x^3}\sqrt {x + y} }}{{8{y^3}\sqrt {x + y} - 1}}\end{aligned}\)

Thus, the value is \(\frac{{dy}}{{dx}} = \frac{{1 - 8{x^3}\sqrt {x + y} }}{{8{y^3}\sqrt {x + y} - 1}}\).