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

Hint: 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.