Solveeit Logo

Question

Question: Differentiate the following function with respect to x. \({{x}^{n}}\tan x\) . (a) \({{x}^{n-1}}\...

Differentiate the following function with respect to x.
xntanx{{x}^{n}}\tan x .
(a) xn1(ntanx+xsecx){{x}^{n-1}}\left( n\tan x+x\sec x \right)
(b) xn1(ntanx+xsec2x){{x}^{n-1}}\left( n\tan x+x{{\sec }^{2}}x \right)
(c) xn1(ntanx+secx){{x}^{n-1}}\left( n\tan x+\sec x \right)
(d) xn1(ntanx+xsec2x){{x}^{n-1}}\left( n\tan x+x{{\sec }^{-2}}x \right)

Explanation

Solution

Hint: In order to crack this problem, we need to know the product rule of differentiation. It is shown as follows, df(x)dx=h(x)dg(x)dx+g(x)dh(x)dx\dfrac{df\left( x \right)}{dx}=h\left( x \right)\dfrac{dg\left( x \right)}{dx}+g\left( x \right)\dfrac{dh\left( x \right)}{dx} . Also, we need to know the individual differentiation formulas for each sun functions like dxndx=nxn1\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}} and dtanxdx=sec2x\dfrac{d\tan x}{dx}={{\sec }^{2}}x .

Complete step-by-step answer:
We aim to find the differentiation of this function with respect to x.
Let the function be named as f(x)=xntanxf\left( x \right)={{x}^{n}}\tan x .
The function f(x)f\left( x \right) contains two sub-functions.
Let g(x)=xng\left( x \right)={{x}^{n}} and h(x)=tanxh\left( x \right)=\tan x be the two sub-functions.
Therefore, f(x)=g(x)×h(x).....................(i)f\left( x \right)=g\left( x \right)\times h\left( x \right).....................\left( i \right) .
To find the differentiation we need to use the product rule of differentiation as the function is the products of two spate functions.
The product rule states as follows,
df(x)dx=h(x)dg(x)dx+g(x)dh(x)dx................(ii)\dfrac{df\left( x \right)}{dx}=h\left( x \right)\dfrac{dg\left( x \right)}{dx}+g\left( x \right)\dfrac{dh\left( x \right)}{dx}................\left( ii \right)
By substituting the values of f(x),g(x),h(x)f\left( x \right),g\left( x \right),h\left( x \right) we get,
df(x)dx=tanxdxndx+xndtanxdx\dfrac{df\left( x \right)}{dx}=\tan x\dfrac{d{{x}^{n}}}{dx}+{{x}^{n}}\dfrac{d\tan x}{dx}
The differentiation xn{{x}^{n}} is dxndx=nxn1..............(iii)\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}..............(iii)
And, the differentiation of tanx\tan x is dtanxdx=sec2x................(iv)\dfrac{d\tan x}{dx}={{\sec }^{2}}x................(iv)
Substituting the values from equation (iii) and (iv) we get,
df(x)dx=tanx(nxn1)+xn(sec2x)\dfrac{df\left( x \right)}{dx}=\tan x\left( n{{x}^{n-1}} \right)+{{x}^{n}}\left( {{\sec }^{2}}x \right)
Simplifying it further we get,
df(x)dx=xn1(ntanx+xsec2x)\dfrac{df\left( x \right)}{dx}={{x}^{n-1}}\left( n\tan x+x{{\sec }^{2}}x \right) .
Hence the option that matches is option (b).

Note: One of the most important points in this problem is to remember the differentiation formulas. The common point of mistake is where we simplify the equation and we take xn1{{x}^{n-1}} coming out of brackets. Therefore, when xn1{{x}^{n-1}} is common, there is xx with sec2x{{\sec }^{2}}x . Also, option (c) is exactly the mistake that is commonly made. As the options are very close we need to be extra careful with this sum.