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The answer I got is in terms of $\sec x$. $$\left.\left[\dfrac{\sec^2x}{x}-\tan x\right] \right/x^2\text{ ?}$$

I simply used the division formula in differentiation to get it.

Not able to get it in terms of $\sin x$ and $\cos x$.

The answer has to be $\dfrac{x-\sin x\cos x}{x^2\cos^2x}$

Michael Hardy
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  • What was your answer? You should include that in the question where people will see it. Even better, show how you got it. That kind of effort tends to get favorable attention even if you made a mistake. – David K Nov 27 '20 at 03:49
  • @DavidK I have written my answer, –  Nov 27 '20 at 03:54
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    The easiest approach is (before diffentiating) convert the tan$(x)$ function into $\frac{\sin(x)}{\cos(x)}.$ Then differentiate. – user2661923 Nov 27 '20 at 04:06
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    There was an issue when you did the quotient rule, but it's been recitified in the answer. You can always convert the other trigonometric ratios to $\sin$ and $\cos$ if you wanted the answer in terms of those functions. – Sarvesh Ravichandran Iyer Nov 27 '20 at 05:41

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You made a mistake in your application of the quotient rule. The derivative is $$ \frac{x\sec^2x - \tan x}{x^2} = \frac{\frac{x}{\cos^2x}-\frac{\sin x}{\cos x}}{x^2} = \frac{\frac{x}{\cos^2x}-\frac{\sin x\cos x}{\cos^2 x}}{x^2} = \frac{x-\sin x\cos x}{x^2\cos^2 x}. $$

DanLewis3264
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