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 .
(a) xn−1(ntanx+xsecx)
(b) xn−1(ntanx+xsec2x)
(c) xn−1(ntanx+secx)
(d) xn−1(ntanx+xsec−2x)
Solution
Hint: In order to crack this problem, we need to know the product rule of differentiation. It is shown as follows, dxdf(x)=h(x)dxdg(x)+g(x)dxdh(x) . Also, we need to know the individual differentiation formulas for each sun functions like dxdxn=nxn−1 and dxdtanx=sec2x .
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)=xntanx .
The function f(x) contains two sub-functions.
Let g(x)=xn and h(x)=tanx be the two sub-functions.
Therefore, f(x)=g(x)×h(x).....................(i) .
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,
dxdf(x)=h(x)dxdg(x)+g(x)dxdh(x)................(ii)
By substituting the values of f(x),g(x),h(x) we get,
dxdf(x)=tanxdxdxn+xndxdtanx
The differentiation xn is dxdxn=nxn−1..............(iii)
And, the differentiation of tanx is dxdtanx=sec2x................(iv)
Substituting the values from equation (iii) and (iv) we get,
dxdf(x)=tanx(nxn−1)+xn(sec2x)
Simplifying it further we get,
dxdf(x)=xn−1(ntanx+xsec2x) .
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 xn−1 coming out of brackets. Therefore, when xn−1 is common, there is x with sec2x . 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.