Question

Find the derivative of $${\displaystyle \frac{1}{x^4\sec x}}$$?
A. $${\displaystyle \frac{\left[4\cos x-x\sin x\right]}{x^5}}$$
B. $${\displaystyle \frac{-\left[x\sin x+4\cos x\right]}{x^5}}$$
C. $${\displaystyle \frac{\left[4\cos x+x\sin x\right]}{x^5}}$$
D. None of these

Answer 1

Hint: We need to find the derivative of $${\displaystyle \frac{1}{x^4\sec x}}$$ with respect to ‘x’. Before differentiating we simplify it by using the definition of secant function. After simplifying we will have equation of the form $$y={\displaystyle \frac{g(x)}{h(x)}}$$, here both $$g(x)$$ and $$h(x)$$ are differentiable. To differentiate it we use the quotient rule, that is $${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{g^{\prime }(x)h(x)-g(x)h^{\prime }(x)}{{(h(x))}^2}}$$.

Complete step by step answer:
Now consider $$y={\displaystyle \frac{1}{x^4\sec x}}$$
We know by the definition of secant function, $$\sec x={\displaystyle \frac{1}{\cos x}}$$. Then
$$y={\displaystyle \frac{1}{x^4\left({\displaystyle \frac{1}{\cos x}}\right)}}$$
$$\Rightarrow y={\displaystyle \frac{\cos x}{x^4}}$$.
Now differentiating with respect to ‘x’
$$\Rightarrow {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{d}{dx}}\left({\displaystyle \frac{\cos x}{x^4}}\right)$$.
Now to differentiate this we use quotient rule,
That is if we have $$f(x)={\displaystyle \frac{g(x)}{h(x)}}$$ and $$h(x)\ne 0$$. Then the quotient rule states that the derivative of $$f(x)$$ is $$f^1(x)={\displaystyle \frac{g^{\prime }(x)h(x)-g(x)h^{\prime }(x)}{{(h(x))}^2}}$$.
If we compare the above equation we have $$g(x)=\cos x$$ and $$h(x)=x^4$$.
$$\Rightarrow {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{d}{dx}}\left({\displaystyle \frac{\cos x}{x^4}}\right)$$
Now applying the quotient rule we have,
$$\Rightarrow {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{\left({\displaystyle \frac{d\left(\cos x\right)}{dx}}.x^4-\cos x.{\displaystyle \frac{d\left(x^4\right)}{dx}}\right)}{{\left(x^4\right)}^2}}$$

We know the differentiation of cosine function that is $${\displaystyle \frac{d\left(\cos x\right)}{dx}}=-\sin x$$ and also using power rule of differentiation $${\displaystyle \frac{d\left(x^4\right)}{dx}}=4x^3$$.
$$\Rightarrow {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{\left(-\sin x.x^4-\cos x.4x^3\right)}{{\left(x^4\right)}^2}}$$
Now taking $$x^3$$common in the numerator we have,
$$\Rightarrow {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{x^3\left(-x.\sin x-4\cos x\right)}{x^8}}$$
Cancelling the ‘x’ terms we have,
$$\Rightarrow {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{\left(-x.\sin x-4\cos x\right)}{x^5}}$$
Taking negative common on the numerator we have,
$$\therefore {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{-\left(x.\sin x+4\cos x\right)}{x^5}}$$.

Hence, the correct answer is option B.

Note: We have several rules of differentiation. We know the power rule of the differentiation that is the definition of $$x^n$$ with respect to ‘x’ is $${\displaystyle \frac{d}{dx}}\left(x^n\right)=n{x}^{n-1}$$. We have used this in the above steps. We also have product rule that is if the function f(x) is the product of two functions, $$y=f(x)g(x)$$ then product rule is given by $$y^1=f^1(x).g(x)+f(x).g^{\prime }(x)$$. We apply these rules depending on the given problem’s.