Question

How do you find the second derivative of $y=\tan \left( x \right)$ ?

Answer 1

In this question, we need to find the second derivative of $y=\tan \left( x \right)$ . For this, we will first take derivative on both sides and find $\dfrac{dy}{dx}$ . We will convert tan(x) to $\dfrac{\sin x}{\cos x}$ and then apply quotient rule to find $\dfrac{dy}{dx}$ . After that, we will find derivative of obtained result using chain rule. It will give us our final answer. We will use formula which are as following:
(I) Quotient rule for two functions u(x) and v(x) is given as $\dfrac{d}{dx}\dfrac{u\left( x \right)}{v\left( x \right)}=\dfrac{v\left( x \right)u'\left( x \right)-v'\left( x \right)u\left( x \right)}{{{\left( v\left( x \right) \right)}^{2}}}$ .
(II) Chain rule is given as $\dfrac{d}{dx}f\left( g\left( x \right) \right)=f'\left( g\left( x \right) \right)\cdot g'\left( x \right)$ .
(III) Derivative of cosx is -sinx and derivative of sinx is cosx.
(IV) secx is reciprocal of cosx.
(V) tanx is equal to $\dfrac{\sin x}{\cos x}$ .
(VI) ${{\cos }^{2}}x+{{\sin }^{2}}x=1$ .

Complete step by step answer:
Here we have to find the second derivative of $y=\tan \left( x \right)$ . We will first find the first derivative of $y=\tan \left( x \right)$ . As we know that, $\tan x=\dfrac{\sin x}{\cos x}$ . So our function becomes $y=\dfrac{\sin x}{\cos x}$ .
$\Rightarrow$ We need to find a derivative of it.
Taking derivative on both sides we get, $\dfrac{dy}{dx}=\dfrac{d}{dx}\left( \dfrac{\sin x}{\cos x} \right)$ .
$\Rightarrow$ Let us use quotient rule here. We know for any two functions u(x) and v(x) derivative by quotient rule is given as, $\dfrac{d}{dx}\dfrac{u\left( x \right)}{v\left( x \right)}=\dfrac{v\left( x \right)u'\left( x \right)-v'\left( x \right)u\left( x \right)}{{{\left( v\left( x \right) \right)}^{2}}}$ .
Here u(x) = sinx and v(x) = cosx.
$\Rightarrow$ We know that, derivative of sinx is cosx and derivative of cosx is -sinx. So u'(x) = cosx and v'(x) = -sinx.
$\Rightarrow$ Coming back to derivative putting values of u(x), v(x), u'(x), v'(x) we get,
$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( \dfrac{\sin x}{\cos x} \right)=\dfrac{\cos x\left( \cos x \right)-\sin x\left( -\sin x \right)}{{{\cos }^{2}}x}\Rightarrow \dfrac{dy}{dx}=\dfrac{{{\cos }^{2}}x+{{\sin }^{2}}x}{{{\cos }^{2}}x}$ .
Now we know that ${{\cos }^{2}}x+{{\sin }^{2}}x$ is equal to 1, therefore let us use it here, we get, $\dfrac{dy}{dx}=\dfrac{1}{{{\cos }^{2}}x}$ .
As we know, secx is reciprocal of cosx, so we write $\dfrac{1}{\cos x}$ as secx. Hence we get $\dfrac{dy}{dx}={{\sec }^{2}}x$ which is the final derivative of $y=\tan \left( x \right)$ .
Now we need to find the second derivative.
Again writing ${{\sec }^{2}}x$ as $\dfrac{1}{{{\cos }^{2}}x}$ we get, $\dfrac{dy}{dx}=\left( \dfrac{1}{{{\cos }^{2}}x} \right)$ .
$\dfrac{1}{{{x}^{2}}}$ can be written as ${{x}^{-2}}$ so here we get $\dfrac{dy}{dx}={{\left( \cos x \right)}^{-2}}$ .
Taking derivatives on both sides we get, $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}{{\left( \cos x \right)}^{-2}}$ .
Applying the chain rule and using $\dfrac{d}{dx}{{\left( x \right)}^{n}}=n{{x}^{n-1}}$ we get,
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-2{{\left( \cos x \right)}^{-2-1}}\cdot \dfrac{d}{dx}\cos x$ .
Using derivative of cosx as -sinx we get, $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-2{{\left( \cos x \right)}^{-3}}\left( -\sin x \right)\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{2\sin x}{{{\left( \cos x \right)}^{3}}}$ .
$\Rightarrow$ We can write ${{\left( \cos x \right)}^{3}}$ as ${{\cos }^{2}}x$ we get $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{2\sin x}{\cos x}\cdot \dfrac{1}{{{\cos }^{2}}x}$.
Using $\tan x=\dfrac{\sin x}{\cos x}\text{ and }\dfrac{1}{{{\cos }^{2}}x}=\sec x$ we get $\dfrac{{{d}^{2}}y}{dx}=2{{\sec }^{2}}x\tan x$ which is our required second derivative.
$\Rightarrow$ Hence the second derivative of $y=\tan \left( x \right)$ is $2{{\sec }^{2}}x\tan x$.

Note:
Students should take care of the signs while solving the derivatives. Note that, they can use direct formulas of derivatives of tanx and secx also which are given as $\dfrac{d}{dx}\tan x=2{{\sec }^{2}}x\text{ and }\dfrac{d}{dx}\sec x=\sec x\tan x$. Do not alter the order in quotient rule.