Question

How do you differentiate $y=\sqrt{1+{{x}^{2}}}?$

Answer 1

Hint: Differentiating a function means classifying it into the parts and the differentiation of \[\sqrt{x}=\dfrac{1}{2\sqrt{x}}\]
As we know that the given equation as $y=\sqrt{1+{{x}^{2}}}$ has a composite function. So, we have to apply chain rule while differentiate.
$ f'(g(x))g'(x)$.

Complete step by step solution:
In the above equation we have to differentiate $y=\sqrt{1+{{x}^{2}}}$
Here we have rewrite the above equation as $y={{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}}$
We had arranged the above equation. Here we see that we have to apply chain rule to differentiate the above equation. So, applying the chain rule to differentiate we get,
$\Rightarrow $\[\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)=f'\left( g\left( x \right) \right).g'\left( x \right)...(i)\]
We have see the chain rule by differentiate
So, now differentiate the value.
$\Rightarrow $$f\left( g\left( x \right) \right)=\dfrac{1}{2}{{\left( 1+{{x}^{2}} \right)}^{-\dfrac{1}{2}}}$
Now,
$\Rightarrow $$g(x)=1+{{x}^{2}}$
When differentiating the above equation we get,
$\Rightarrow $$g'(x)=2x$
Here, when we differentiate the constant value we have zero.
Now, substituting the above value in equation $(i)$ we get,
$\Rightarrow $$\dfrac{1}{2}{{\left( 1+{{x}^{2}} \right)}^{-\dfrac{1}{2}}}.2x=x{{\left( 1+{{x}^{2}} \right)}^{-\dfrac{1}{2}}}$
When differentiate, we have
$\Rightarrow $$\dfrac{dy}{dx}=\dfrac{x}{{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}}}$
We can write it as follow
$\Rightarrow $$\dfrac{dy}{dx}=\dfrac{x}{\sqrt{1+{{x}^{2}}}}$

Note: The general form of derivation for applying chain rule.
We use chain rule when the function is of composite function.
Here in chain rule we assume derivatives of the outside element with inside element are in the same.
Let see the general equation
$\Rightarrow $$f'\left( g\left( x \right) \right).g'\left( x \right)$
For example,
We have to find the derivative of \[\sin \left( 5x \right)\] as we know the above element is a composite function.
We have to apply chain rule.
First $f(x)=\sin x$
Now, differentiating above equation we get,
$\Rightarrow $$f'(x)=\cos x$
Now, differentiate $\sin x$ we get $\cos x$
$\Rightarrow $$g(x)=5x$
Now differentiating above value we have
$\Rightarrow $$g'(x)=5$
When we differentiate $5x$ we get $5$
$\Rightarrow $$\cos \left( 5x \right)5=5\cos \left( 5x \right)$
When we are going to differentiate a constant term we get value as zero while differentiate we have to follow the differentiate rule.
When differentiating the function which has one function included in another function then that function should be operated separately.