# Verify that the given function(explicit or implicit)is a solution of the corresponding differential equation$y=\sqrt{1+x^2}\qquad:\;y'\;=\frac{xy}{1+x^2}$

## 1 Answer

Toolbox:
• Differentiation of $\sqrt {1+ x^2}$ is $\large\frac{2x}{2\sqrt{ (1+x^2)} } =\large\frac{ x}{1+x^2}$
Step 1:
Given $y = \sqrt{1 + x^2}$
Differentiating on both sides we get
$y' =\large\frac{ 1}{2}$$(1 + x^2)^{\Large\frac{-1}{2}}$$ \times 2x$
$y' = \large\frac{x}{\sqrt{1 + x^2}}$
Step 2:
Multiply the numerator and denominator by $\sqrt{1+x^2}$ we get
$y' =\large\frac{x\sqrt{ (1+x^2)}}{(1+x^2)} = \frac{xy}{1+x^2}$
Hence the answer is a solution to the given function.
answered Aug 19, 2013

1 answer

1 answer