Question

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

Answer 1

We need to find the derivative of $\dfrac{1}{{{x^4}\sec x}}$ 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 $y = \dfrac{{g(x)}}{{h(x)}}$, here both $g(x)$ and $h(x)$ are differentiable. To differentiate it we use the quotient rule, that is $\dfrac{{dy}}{{dx}} = \dfrac{{g'(x)h(x) - g(x)h'(x)}}{{{{(h(x))}^2}}}$.

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

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

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 $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$. 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 ${y^1} = {f^1}(x).g(x) + f(x).g'(x)$. We apply these rules depending on the given problem’s.