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Question: What is the derivative of the inverse tangent of 2x?...

What is the derivative of the inverse tangent of 2x?

Explanation

Solution

In this problem, we have to find the derivative of the inverse tangent of 2x. Here we have to convert the given statement into its mathematical form. After forming an equation, we can differentiate them on both sides with respect to x to get the derivative of the given problem.

Complete step-by-step solution:
Here we have to find the derivative of the inverse tangent of 2x.
We can now write the given question as,
y=tan12xy={{\tan }^{-1}}2x
We can now write the above step as,
tany=2x\Rightarrow \tan y=2x ……. (1)
we can now differentiate the above step on both sides with respect to x, we get
ddx(tany)=ddx(2x)\Rightarrow \dfrac{d}{dx}\left( \tan y \right)=\dfrac{d}{dx}\left( 2x \right)
We can now differentiate the above step as we differentiate, tanxdx=sec2x\tan xdx={{\sec }^{2}}x and 2xdx=22xdx=2, we get
sec2y(dydx)=2\Rightarrow {{\sec }^{2}}y\left( \dfrac{dy}{dx} \right)=2
We can now write the above step as,
dydx=2sec2y\Rightarrow \dfrac{dy}{dx}=\dfrac{2}{{{\sec }^{2}}y}
We know that sec2x=1+tan2x{{\sec }^{2}}x=1+{{\tan }^{2}}x, we can now apply this in the above step, we get
dydx=21+tan2y\Rightarrow \dfrac{dy}{dx}=\dfrac{2}{1+{{\tan }^{2}}y} …… (2)
Now from (1), we can write as

& \Rightarrow {{\tan }^{2}}y={{\left( 2x \right)}^{2}} \\\ & \Rightarrow {{\tan }^{2}}y=4{{x}^{2}}.....(3) \\\ \end{aligned}$$ We can now substitute (3) in (2), we get $$\Rightarrow \dfrac{dy}{dx}=\dfrac{2}{1+4{{x}^{2}}}$$ **Therefore, the derivative of inverse tangent of 2x is $$\dfrac{dy}{dx}=\dfrac{2}{1+4{{x}^{2}}}$$** **Note:** We can also find the derivative of the formula method. We know that the given equation is, $$y={{\tan }^{-1}}2x$$ We can now differentiate it, we get $$\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{\tan }^{-1}}\left( 2x \right) \right)$$ …….. (1) We know that, $$\dfrac{d}{dx}{{\tan }^{-1}}x=\dfrac{1}{1+{{x}^{2}}}$$. We can now apply the above formula in (1), we get $$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{1+{{\left( 2x \right)}^{2}}}\dfrac{d}{dx}\left( 2x \right)$$ We can now differentiate the above step, we get $$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{1+{{\left( 2x \right)}^{2}}}\times 2$$ We can now simplify the above step, we get $$\Rightarrow \dfrac{dy}{dx}=\dfrac{2}{1+4{{x}^{2}}}$$ Therefore, the derivative of inverse tangent of 2x is $$\dfrac{dy}{dx}=\dfrac{2}{1+4{{x}^{2}}}$$