Question

How do you differentiate y = $x\tan \left( x \right)$ ?

Answer 1

Hint: Here in this question, we will be applying differentiation formulae for trigonometric functions. For tan(x), the differentiation formula is ${{\sec }^{2}}x$. Also, we have to apply product rules as x and tan(x) are in product form. The product rule of differentiation is:
$\Rightarrow \dfrac{d}{dx}(fg)=f'g+fg'$
 Where f and g are the functions of ‘x’ and differentiated with respect to ‘x’.

Complete step by step answer:
Let’s discuss the question now.
Differentiation is simply the method of obtaining the derivative of a function. There are various rules for finding the derivative of function(s). Let f and g as functions of ‘x’ and differentiate with respect to ‘x’ and learn how to apply rules on the functions:
Power rule: $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$
Derivative of a constant, a: $\dfrac{d}{dx}(a)=0$
Derivative of a constant multiplied with function f: $\dfrac{d}{dx}(a.f)=af'$
Sum rule: $\dfrac{d}{dx}(f\pm g)=f'\pm g'$
Product rule: $\dfrac{d}{dx}(fg)=f'g+fg'$
Quotient rule: $\dfrac{d}{dx}\left( \dfrac{f}{g} \right)=\dfrac{gf'-fg'}{{{g}^{2}}}$
Apart from these, we need to learn differentiation formulae for trigonometric functions as well.
$\begin{align}
  & \Rightarrow \dfrac{d}{dx}\left( \sin x \right)=\cos x \\
 & \Rightarrow \dfrac{d}{dx}\left( \cos x \right)=-\sin x \\
 & \Rightarrow \dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x \\
 & \Rightarrow \dfrac{d}{dx}\left( \cot x \right)=-\cos e{{c}^{2}}x \\
 & \Rightarrow \dfrac{d}{dx}\left( \sec x \right)=\sec x\tan x \\
 & \Rightarrow \dfrac{d}{dx}\left( \cos ecx \right)=-\cos ecx\cot x \\
\end{align}$
As we know that tanx = $\dfrac{\sin x}{\cos x}$. so if we find $\dfrac{d}{dx}\left( \dfrac{\sin x}{\cos x} \right)$ we actually find the derivative of tanx.
Let’s see how we can derive the formula.
To find $\dfrac{d}{dx}\left( \dfrac{\sin x}{\cos x} \right)$, we have to apply quotient rule: $\dfrac{d}{dx}(\dfrac{f}{g})=\dfrac{gf'-fg'}{{{g}^{2}}}$.
By applying the rule, we get:
$\Rightarrow \dfrac{d}{dx}\left( \tan x \right)=\dfrac{d}{dx}\left( \dfrac{\sin x}{\cos x} \right)$
$\Rightarrow \dfrac{d}{dx}\left( \tan x \right)=\dfrac{\cos x.\dfrac{d}{dx}\left( \sin x \right)-\dfrac{d}{dx}\left( \cos x \right).\sin x}{{{\cos }^{2}}x}$
From above formulae, $\dfrac{d}{dx}\left( \sin x \right)=\cos x$ and $\dfrac{d}{dx}\left( \cos x \right)=-\sin x$. Insert the values in the equation, we get:
$\Rightarrow \dfrac{d}{dx}\left( \tan x \right)=\dfrac{\cos x.\left( \cos x \right)+\left( \sin x \right).\sin x}{{{\cos }^{2}}x}$
Now, solve further:
$\Rightarrow \dfrac{d}{dx}\left( \tan x \right)=\dfrac{{{\cos }^{2}}x+{{\sin }^{2}}x}{{{\cos }^{2}}x}$
We already know that ${{\cos }^{2}}x+{{\sin }^{2}}x=1$. So by putting the value, we get:
$\Rightarrow \dfrac{d}{dx}\left( \tan x \right)=\dfrac{1}{{{\cos }^{2}}x}$
We also know that $\dfrac{1}{cox}=\sec x$, so:
$\Rightarrow \dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x$
Now write the equation given in question:
$\Rightarrow $y = $x\tan \left( x \right)$
Now, we can see that x and tan(x) are in product. So we will apply product rule:
$\Rightarrow \dfrac{d}{dx}(fg)=f'g+fg'$
By applying the rule, we get:
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}(x).\tan (x)+x\dfrac{d}{dx}(\tan x)$
As we know that $\dfrac{d}{dx}(x)=1$ and $\dfrac{d}{dx}(\tan x)={{\sec }^{2}}x$. Putting the values we get:
$\Rightarrow \dfrac{dy}{dx}=1.\tan (x)+x.{{\sec }^{2}}x$
${{\sec }^{2}}x$ is nothing but $\dfrac{1}{{{\cos }^{2}}x}$. Now solve further:
 $\Rightarrow \dfrac{dy}{dx}=\tan (x)+\dfrac{x}{{{\cos }^{2}}x}$
So, finally we get the derivative of y = $x\tan \left( x \right)$.

Note: The rules are to be followed while solving any question related to differentiation. Derivatives of all trigonometric functions should be on tips so that you don’t need to obtain the derivatives for derived functions. You can also directly put the value of derivative of tan(x) and solve. But if the question comes for more weightage, you can obtain the derivative of tan(x) as well.