Question

If an inverse trigonometric function is given by $y={{\tan }^{-1}}\left( \dfrac{x-\sqrt{x}}{1+{{x}^{\dfrac{3}{2}}}} \right)$ , then find out the value of ${{\dfrac{dy}{dx}}_{x=1}}$ .

Answer 1

Hint:We can rearrange and substitute $x=\tan A$ and ${{x}^{\dfrac{1}{2}}}=\tan B$. Then, we can use the chain rule of differentiation for $f(g(x))$ given by ${{f}^{'}}(g(x))\times {{g}^{'}}(x)$ and then solve it.

Complete step-by-step solution -

So we have been given that $y={{\tan }^{-1}}\left( \dfrac{x-\sqrt{x}}{1+{{x}^{\dfrac{3}{2}}}} \right)$.
So we can see that we can make the above equation in simpler form. If we want to differentiate in simpler form we should cancel the ${{\tan }^{-1}}$ . So for removing ${{\tan }^{-1}}$ we have to substitute $x$ and ${{x}^{\dfrac{1}{2}}}$ .
So simplifying it we get
$y={{\tan }^{-1}}\left( \dfrac{x-{{x}^{\dfrac{1}{2}}}}{1+x{{x}^{\dfrac{1}{2}}}} \right)$ ………… (1)
So let us consider $x=\tan A$ and ${{x}^{\dfrac{1}{2}}}=\tan B$ . Now substituting $x$ and ${{x}^{\dfrac{1}{2}}}$ by $\tan A$ and $\tan B$ we get,
$y={{\tan }^{-1}}\left( \dfrac{\tan A-\tan B}{1+\tan A\tan B} \right)$ ……….. (2)
So we know the identity $\tan (A-B)$ ,
$\tan (A-B)=\left( \dfrac{\tan A-\tan B}{1-\tan A\tan B} \right)$ …… (3)
Now substituting (3) in (2), we get,
$\begin{aligned} & y={{\tan }^{-1}}\left( \dfrac{\tan A-\tan B}{1+\tan A\tan B} \right) \\\ & y={{\tan }^{-1}}(\tan (A-B)) \\\ \end{aligned}$
So here ${{\tan }^{-1}}$ and $\tan$ get cancelled, we get $y=A-B$. So now resubstituting we get,
$y={{\tan }^{-1}}x-{{\tan }^{-1}}({{x}^{\dfrac{1}{2}}})$ …………. ($A={{\tan }^{-1}}x$and$B={{\tan }^{-1}}({{x}^{\dfrac{1}{2}}})$)
Now we know if $y={{\tan }^{-1}}x$, then the differentiation becomes,
$\dfrac{dy}{dx}=\dfrac{1}{1+{{x}^{2}}}$
Now we want to find for $y={{\tan }^{-1}}x-{{\tan }^{-1}}({{x}^{\dfrac{1}{2}}})$ . So for differentiating, we have to use chain rule in the ${{\tan }^{-1}}\left( {{x}^{\dfrac{1}{2}}} \right)$ .
So the chain rule tells us how to find the derivative of a composite function. The chain rule states that the derivative of $f(g(x))$ is ${{f}^{'}}(g(x))\times {{g}^{'}}(x)$ . In other words, it helps us differentiate composite functions.
So differentiating and applying chain rule for ${{\tan }^{-1}}\left( {{x}^{\dfrac{1}{2}}} \right)$ we get,
$\begin{aligned} & \dfrac{dy}{dx}=\dfrac{1}{1+{{x}^{2}}}-\dfrac{1}{1+{{({{x}^{\dfrac{1}{2}}})}^{2}}}\dfrac{d({{x}^{\dfrac{1}{2}}})}{dx} \\\ & \dfrac{dy}{dx}=\dfrac{1}{1+{{x}^{2}}}-\dfrac{1}{1+{{({{x}^{\dfrac{1}{2}}})}^{2}}}\left( \dfrac{1}{2\sqrt{x}} \right) \\\ \end{aligned}$
Simplifying we get,
$\dfrac{dy}{dx}=\dfrac{1}{1+{{x}^{2}}}-\dfrac{1}{2(1+x)\sqrt{x}}$ ……… (4)
Now we want to differentiate at $x=1$ , ……………………. (given that differentiate at $x=1$)
So substituting $x=1$ in (4), we get,
$\begin{aligned} & {{\dfrac{dy}{dx}}_{x=1}}=\dfrac{1}{1+{{1}^{2}}}-\dfrac{1}{2(1+1)\sqrt{1}} \\\ & {{\dfrac{dy}{dx}}_{x=1}}=\dfrac{1}{1+1}-\dfrac{1}{2(2)} \\\ & {{\dfrac{dy}{dx}}_{x=1}}=\dfrac{1}{2}-\dfrac{1}{4} \\\ \end{aligned}$
So to solve above we must take LCM
$\begin{aligned} & {{\dfrac{dy}{dx}}_{x=1}}=\dfrac{2-1}{4} \\\ & {{\dfrac{dy}{dx}}_{x=1}}=\dfrac{1}{4} \\\ \end{aligned}$
${{\dfrac{dy}{dx}}_{x=1}}=\dfrac{1}{4}$
So we get ${{\dfrac{dy}{dx}}_{x=1}}$ as $\dfrac{1}{4}$ .

Note: You should be familiar with the properties or identity such as $\tan (A-B)=\left( \dfrac{\tan A-\tan B}{1-\tan A\tan B} \right)$ . While differentiating $\dfrac{dy}{dx}$ be careful that you do not miss any of the value for differentiating. Take utmost care for differentiating ${{\tan }^{-1}}$ since it is complicated. While substituting $x=1$ be careful that you do not miss it anywhere.