Question

If \[x = a\left( {\cos t + \log \tan \dfrac{t}{2}} \right)\]and \[y = a\sin t\], then what is the value of\[\dfrac{{dy}}{{dx}}\]?
A.\[\tan t\]
B.\[ - \tan t\]
C.\[\cot t\]
D.\[ - \cot t\]

Answer 1

Hint: First we will derivative \[x = a\left( {\cos t + \log \tan \dfrac{t}{2}} \right)\] with respect to \[t\] using the formulas \[\dfrac{d}{{dx}}\cos x = - \sin x\], \[\dfrac{d}{{dx}}\log x = \dfrac{1}{x}\] and \[\dfrac{d}{{dx}}\tan x = {\sec ^2}x\]. Then we will simplify the equation. Again we will derivative \[y = a\sin t\] with respect to \[t\] using the formula \[\dfrac{d}{{dx}}\sin x = \cos x\]. After that we will apply the formula \[\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}}\] to get the required answer.

Formula used :
We will use the derivative formulas
\[\dfrac{d}{{dx}}\cos x = - \sin x\], \[\dfrac{d}{{dx}}\sin x = \cos x\],\[\dfrac{d}{{dx}}\log x = \dfrac{1}{x}\],\[\dfrac{d}{{dx}}\tan x = {\sec ^2}x\] ,\[\dfrac{d}{{dx}}x = 1\]and we will also use the chain rule, and also trigonometric formulas, \[\tan x = \dfrac{{\sin x}}{{\cos x}}\]and \[\sec x = \dfrac{1}{{\cos x}}\], \[\sin x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}\],\[1 - {\sin ^2}t = {\cos ^2}t\]

Complete Step-by- Step Solution:
Given \[x = a\left( {\cos t + \log \tan \dfrac{t}{2}} \right)\], and \[y = a\sin t\],
Now we will take the \[x\] value and differentiate with respect to \[t\] on both sides,
\[ \Rightarrow x = a\left( {\cos t + \log \tan \dfrac{t}{2}} \right)\]
Now we will differentiate with respect to\[t\] on both sides,
\[ \Rightarrow \dfrac{{dx}}{{dt}} = \dfrac{d}{{dt}}\left( {a\left( {\cos t + \log \tan \dfrac{t}{2}} \right)} \right)\],
Now will distribute the differentiation,
\[ \Rightarrow \dfrac{{dx}}{{dt}} = a\left( {\dfrac{d}{{dt}}\left( {\cos t} \right) + \dfrac{d}{{dt}}\log \tan \dfrac{t}{2}} \right)\]
Now we will apply derivative formulas i.e.,\[\dfrac{d}{{dx}}\cos x = - \sin x\],\[\dfrac{d}{{dx}}\log x = \dfrac{1}{x}\] and also applying chain rule, then we will get,
\[ \Rightarrow \dfrac{{dx}}{{dt}} = a\left( { - \sin t + \dfrac{1}{{\tan \dfrac{t}{2}}}\dfrac{d}{{dt}}\tan \dfrac{t}{2}} \right)\]
Now again we will use the derivative formula \[\dfrac{d}{{dx}}\tan x = {\sec ^2}x\] and also chain rule,
\[ \Rightarrow \dfrac{{dx}}{{dt}} = a\left( { - \sin t + \dfrac{1}{{\tan \dfrac{t}{2}}}\left( {{{\sec }^2}\dfrac{t}{2}} \right)\dfrac{d}{{dt}}\left( {\dfrac{t}{2}} \right)} \right)\]
Now again we will use the derivative formula i.e., \[\dfrac{d}{{dx}}x = 1\], we will get,
\[ \Rightarrow \dfrac{{dx}}{{dt}} = a\left( { - \sin t + \dfrac{1}{{\tan \dfrac{t}{2}}}\left( {{{\sec }^2}\dfrac{t}{2}} \right)\left( {\dfrac{1}{2}} \right)} \right)\]
Now we will simplify the expression using trigonometric formulas, i.e, \[\tan x = \dfrac{{\sin x}}{{\cos x}}\]and \[\sec x = \dfrac{1}{{\cos x}}\], we will get,
\[ \Rightarrow \dfrac{{dx}}{{dt}} = a\left( { - \sin t + \dfrac{1}{{\dfrac{{\sin \dfrac{t}{2}}}{{\cos \dfrac{t}{2}}}}}\left( {\dfrac{1}{{{{\cos }^2}\dfrac{t}{2}}}} \right)\left( {\dfrac{1}{2}} \right)} \right)\]
Now we will further simplify, we will get,
\[ \Rightarrow \dfrac{{dx}}{{dt}} = a\left( { - \sin t + \dfrac{{\cos \dfrac{t}{2}}}{{\sin \dfrac{t}{2}}}\left( {\dfrac{1}{{{{\cos }^2}\dfrac{t}{2}}}} \right)\left( {\dfrac{1}{2}} \right)} \right)\]
Now we will eliminate the like terms we will get,
\[ \Rightarrow \dfrac{{dx}}{{dt}} = a\left( { - \sin t + \dfrac{1}{{2\sin \dfrac{t}{2}\cos \dfrac{t}{2}}}} \right)\]
Now we will use the trigonometric formula i.e., \[\sin x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}\], then we will get,
\[ \Rightarrow \dfrac{{dx}}{{dt}} = a\left( { - \sin t + \dfrac{1}{{\sin t}}} \right)\]
Now we will simplify further, then we will get,
\[ \Rightarrow \dfrac{{dx}}{{dt}} = a\left( {\dfrac{{1 - {{\sin }^2}t}}{{\sin t}}} \right)\]
Now we will again use trigonometric formula, I.e., \[1 - {\sin ^2}t = {\cos ^2}t\], then we will get,
\[ \Rightarrow \dfrac{{dx}}{{dt}} = a\left( {\dfrac{{{{\cos }^2}t}}{{\sin t}}} \right)\]
Now we will take the \[y\] value and differentiate with respect to \[t\] on both sides,
\[y = a\sin t\]
Now differentiate with respect to \[t\] on both sides,
\[\dfrac{{dy}}{{dt}} = \dfrac{d}{{dt}}a\sin t\]
Now we will take out the constant term, we will get,
\[\dfrac{{dy}}{{dt}} = a\dfrac{d}{{dt}}\sin t\]
Now we will use the derivative formula, i.e., \[\dfrac{d}{{dx}}\sin x = \cos x\], we will get,
\[\dfrac{{dy}}{{dt}} = a\cos t\]
Now we will divide both the results, we will get,
\[ \Rightarrow \dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}} = \dfrac{{a\cos t}}{{a\left( {\dfrac{{{{\cos }^2}t}}{{\sin t}}} \right)}}\]
Now we will eliminate the like terms, we will get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\sin t}}{{\cos t}}\]
Now we will use the trigonometric formula i.e.,\[\tan x = \dfrac{{\sin x}}{{\cos x}}\], we will get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \tan t\].
The correct option is A.

 Note: Students often do a common mistake. They used the formula \[\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{dt}} \cdot \dfrac{{dx}}{{dt}}\] which is an incorrect formula. The correct formula is \[\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}}\].