The derivative of 1 by 1+x^2 is equal to -2x/(1+x^2)^2. In this post, we find the derivative of 1/(1+x^2) using the chain rule and substitution method.
Derivative of 1/(1+x^2)
Let us find the derivative of 1/(1+x^2) by the substitution method and the product rule of derivatives. Suppose that
$u=\dfrac{1}{1+x^2}$ …(I)
We will need to find the value of $\dfrac{du}{dx}$.
Step 1: Form (I), we have that
$u(1+x^2)=1$ …(II)
Step 2: Note that $u(1+x^2)$ is a product of two functions. So we will use the product rule of derivatives. Now, differentiating (II) with respect to $x$, we get that
$\dfrac{du}{dx}(1+x^2)+u(0+2x)=0$
$\Rightarrow \dfrac{du}{dx}(1+x^2)+u(0+2x)=0$
$\Rightarrow \dfrac{du}{dx}(1+x^2)+2ux=0$
$\Rightarrow \dfrac{du}{dx}(1+x^2)=-2ux$
$\Rightarrow \dfrac{du}{dx}$
$=-\dfrac{2x}{1+x^2} \cdot u$
Step 3: We now put the value of $u$. By doing so the above is equal to
$\dfrac{du}{dx}$ $=-\dfrac{2x}{1+x^2} \cdot \dfrac{1}{1+x^2}$ as we have $u=\dfrac{1}{1+x^2}$
$\Rightarrow \dfrac{du}{dx}$ $=-\dfrac{2x}{(1+x^2)^2}$
Thus, we obtain that the derivative of $\dfrac{1}{1+x^2}$ is equal to $-\dfrac{2x}{(1+x^2)^2}$.
Derivative of 1/(1+x^2) by Chain Rule
Let $t=1+x^2$.
Then $\dfrac{dt}{dx}=2x$.
Now, we have that
$\dfrac{1}{1+x^2}=\dfrac{1}{t^2}$
Differentiating both sides with respect to $x$, we obtain that
$\dfrac{d}{dx}(\dfrac{1}{1+x^2})$ $=\dfrac{d}{dx}(\dfrac{1}{t^2})$
$=\dfrac{d}{dt}(\dfrac{1}{t}) \cdot \dfrac{dt}{dx}$ by the chain rule of derivatives.
$=\dfrac{d}{dt}(t^{-1}) \cdot 2x$ as dt/dx=2x.
$=-1 \cdot t^{-1-1} \cdot 2x$ by the power rule of derivatives $\dfrac{d}{dx}(x^n)=nx^{n-1}$
$=-2x \cdot \dfrac{1}{t^2}$
$=-2x \cdot \dfrac{1}{(1+x^2)^2}$ as t=1+x^2
$=-\dfrac{2x}{(1+x^2)^2}$
Thus, the derivative of 1/(1+x^2) is equal to -2x/(1+x^2)^2, obtained by the chain rule and the power rule of derivatives.
Also Read:
Derivative of root x + 1/root x
FAQs
Q1: What is the derivative of 1/1+x^2?
Answer: The derivative of 1/(1+x^2) is equal to -2x/(1+x^2)^2.