Question

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

Answer 1

Hint: 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.