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I have no idea what I have done wrong. Please criticise.

Differentiate $y = \frac{1 - \sec x}{\tan x}$.

\begin{align*} \frac{d}{dx} \left[ \frac{1 - \sec x}{\tan x} \right] &= \frac{\tan x \frac{d}{dx} [ 1 - \sec x ] - (1 - \sec x) \frac{d}{dx} \tan x }{(\tan x)^2} \\ &= \frac{\tan x (- \sec x \tan x) - [(1 - \sec x) (-\csc^2 x)]}{(\tan x)^2} \\ &= \frac{-\tan^2 x \sec x + \csc^2 x - \sec x \csc^2 x}{(\tan x)^2} \end{align*}

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    $\begingroup$ Here is a MathJax tutorial. $\endgroup$ – Aaron Maroja Mar 13 '15 at 21:13
  • $\begingroup$ The question appears to ask for the derivative of $1 - \frac{\sec x}{\tan x}$, but in your work you start with $\frac{1 - \sec x}{\tan x}$. $\endgroup$ – dalastboss Mar 13 '15 at 21:17
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Hint: $$ \dfrac{d}{dx} \tan x= \sec^2 x $$ Here you was wrong.

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  • $\begingroup$ Well now I have -tan^2 x sec x - sec^2 x + sec^3x /(tan x)^2 $\endgroup$ – Cetshwayo Mar 13 '15 at 21:29
  • $\begingroup$ yes. But Use MatJax !!!! Look at the link in Aaron comment. $\endgroup$ – Emilio Novati Mar 13 '15 at 21:37
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The derivative of $\tan x$ is $\sec^2 x$. You wrote that it is $-\csc^2x$, in the second line. You can proceed like: $$\frac{\tan x(-\sec x \tan x) - (1-\sec x)\sec^2x}{\tan^2x} = -\sec x - \frac{(1-\sec x)}{\sin^2 x} = \cdots$$

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  • $\begingroup$ Can you explain the second part of your explanation? $\endgroup$ – Cetshwayo Mar 13 '15 at 21:31
  • $\begingroup$ I simplified both $\tan x$ in the numerator with $\tan^2x$ in the denominator, for the first term. For the second term I used: $$\tan^2(x) = \frac{\sin^2x}{\cos^2x} \implies \frac{1}{\tan^2x}=\frac{\cos^2x}{\sin^2x} \implies \frac{\sec^2x}{\tan^2x}=\frac{\sec^2x\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}.$$ $\endgroup$ – Ivo Terek Mar 13 '15 at 21:51
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This is equivalent to $1-\csc x$

$y'=\cot x \csc x$

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    $\begingroup$ Yes it is! +1 ... makes taking the derivative trivial in comparison to using the product rule. Perhaps the OP was checking application of the product rule. $\endgroup$ – Mark Viola Mar 13 '15 at 21:28
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    $\begingroup$ I was going to do the question the same way until I saw this answer - I think there might have been an edit in bracketing in the question which makes the function $\cot x - \csc x$ - but the general principle - simplify before differentiating - is a good one to bear in mind. $\endgroup$ – Mark Bennet Mar 13 '15 at 21:45

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