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Question

Question: How do you differentiate \(\dfrac{x}{{\cos x}}\)?...

How do you differentiate xcosx\dfrac{x}{{\cos x}}?

Explanation

Solution

This problem deals with differentiating the given function using quotient rule. The quotient rule is a formula for taking the derivative of a quotient of two functions. The formula states that to find the derivative of f(x)f(x) divided by g(x)g(x). This is done by the ratio of the g(x)g(x) into the derivative of f(x)f(x), then from the product, we must subtract the product of f(x)f(x) times the derivative of g(x)g(x). The quotient rule is shown below:
ddx(f(x)g(x))=g(x)ddxf(x)f(x)ddxg(x)(g(x))2\Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{f(x)}}{{g(x)}}} \right) = \dfrac{{g(x)\dfrac{d}{{dx}}f(x) - f(x)\dfrac{d}{{dx}}g(x)}}{{{{\left( {g(x)} \right)}^2}}}

Complete step-by-step solution:
The given function is a ratio of the term xx and cosine trigonometric function of xx.
The function is considered as given below:
xcosx\Rightarrow \dfrac{x}{{\cos x}}
Now applying the quotient rule to given function as shown below:
ddx(xcosx)\Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{x}{{\cos x}}} \right)
xddx(cosx)cosxddx(x)(cosx)2\Rightarrow \dfrac{{x\dfrac{d}{{dx}}\left( {\cos x} \right) - \cos x\dfrac{d}{{dx}}\left( x \right)}}{{{{\left( {\cos x} \right)}^2}}}
We know that the derivative of cosx\cos x is sinx - \sin x, as shown:
x(sinx)cosx(1)(cosx)2\Rightarrow \dfrac{{x\left( { - \sin x} \right) - \cos x\left( 1 \right)}}{{{{\left( {\cos x} \right)}^2}}}
xsinxcosxcos2x\Rightarrow \dfrac{{ - x\sin x - \cos x}}{{{{\cos }^2}x}}
Now splitting the fractions, as shown below:
xsinxcos2xcosxcos2x\Rightarrow \dfrac{{ - x\sin x}}{{{{\cos }^2}x}} - \dfrac{{\cos x}}{{{{\cos }^2}x}}
xsinxcosx1cosx1cosx\Rightarrow \dfrac{{ - x\sin x}}{{\cos x}} \cdot \dfrac{1}{{\cos x}} - \dfrac{1}{{\cos x}}
As we know that the ratio of sinx\sin x and cosx\cos x is equal to tanx\tan x, and the reciprocal of cosx\cos x is secx\sec x.
xtanxsecxsecx\Rightarrow - x\tan x\sec x - \sec x
secx(1+xtanx)\Rightarrow - \sec x\left( {1 + x\tan x} \right)

The differentiation of xcosx\dfrac{x}{{\cos x}} is secx(1+xtanx) - \sec x\left( {1 + x\tan x} \right).

Note: Please note that differential calculus is a subfield of calculus that studies the rates at which quantities change. It is one of the two traditional divisions of calculus. The derivative of a function of a real variable measures the sensitivity to change of the function value with respect to a change in its argument. Derivatives are a fundamental tool of calculus.
There are many applications of differential calculus, with which we can find the rate of change of one quantity with respect to another. The rate of change of xx with respect to yy is expressed as dxdy\dfrac{{dx}}{{dy}}.