GSEB Class 12 Maths Solutions Chapter 5 Continuity and Differentiability Exercise 5.6

Question 1. If \( x = 2at^2, y = at^4 \) then find \( \frac{dy}{dx} \).
Answer: We are given the parametric equations:
\( x = 2at^2 \)
\( y = at^4 \)
First, we find the derivatives of \( x \) and \( y \) with respect to \( t \):
\( \frac{dx}{dt} = \frac{d}{dt}(2at^2) = 2a \cdot 2t = 4at \)
\( \frac{dy}{dt} = \frac{d}{dt}(at^4) = a \cdot 4t^3 = 4at^3 \)
Now, we can find \( \frac{dy}{dx} \) using the chain rule:
\( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4at^3}{4at} \)
\( \implies \frac{dy}{dx} = t^2 \)
In simple words: We find how fast \( x \) and \( y \) change with \( t \), then we divide those rates to get how fast \( y \) changes with \( x \), which gives \( t^2 \).

Exam Tip: Remember to differentiate both \( x \) and \( y \) with respect to the parameter (here, \( t \)) separately before forming the ratio to find \( \frac{dy}{dx} \).

Question 2. If \( x = a \cos \theta, y = b \cos \theta \) then find \( \frac{dy}{dx} \).
Answer: We have the parametric equations:
\( x = a \cos \theta \)
\( y = b \cos \theta \)
First, we differentiate \( x \) and \( y \) with respect to \( \theta \):
\( \frac{dx}{d\theta} = \frac{d}{d\theta}(a \cos \theta) = -a \sin \theta \)
\( \frac{dy}{d\theta} = \frac{d}{d\theta}(b \cos \theta) = -b \sin \theta \)
Now, we calculate \( \frac{dy}{dx} \) using the chain rule:
\( \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{-b \sin \theta}{-a \sin \theta} \)
\( \implies \frac{dy}{dx} = \frac{b}{a} \)
In simple words: We calculate how \( x \) and \( y \) change as \( \theta \) changes. Then we divide those change rates to determine how \( y \) changes with \( x \), resulting in \( \frac{b}{a} \).

Exam Tip: Pay attention to the signs of the derivatives of trigonometric functions. The sine terms cancel out, making the result a simple ratio of constants.

Question 3. If \( x = \sin t, y = \cos 2t \) then find \( \frac{dy}{dx} \).
Answer: We are given the equations:
\( x = \sin t \)
\( y = \cos 2t \)
First, we differentiate \( x \) and \( y \) with respect to \( t \):
\( \frac{dx}{dt} = \frac{d}{dt}(\sin t) = \cos t \)
\( \frac{dy}{dt} = \frac{d}{dt}(\cos 2t) = -2 \sin 2t \)
We know that \( \sin 2t = 2 \sin t \cos t \). So, \( \frac{dy}{dt} = -2(2 \sin t \cos t) = -4 \sin t \cos t \).
Now, we find \( \frac{dy}{dx} \) using the chain rule:
\( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-4 \sin t \cos t}{\cos t} \)
\( \implies \frac{dy}{dx} = -4 \sin t \)
In simple words: We find the rates of change for \( x \) and \( y \) concerning \( t \). After that, we divide these rates to get the rate of change for \( y \) concerning \( x \), simplifying to \( -4 \sin t \).

Exam Tip: When differentiating trigonometric functions involving multiple angles (like \( \cos 2t \)), remember to apply the chain rule correctly. Using identities like \( \sin 2t = 2 \sin t \cos t \) can simplify the final expression.

Question 4. If \( x = 4t, y = \frac{4}{t} \) then find \( \frac{dy}{dx} \).
Answer: We have the given equations:
\( x = 4t \)
\( y = \frac{4}{t} = 4t^{-1} \)
First, we differentiate \( x \) and \( y \) with respect to \( t \):
\( \frac{dx}{dt} = \frac{d}{dt}(4t) = 4 \)
\( \frac{dy}{dt} = \frac{d}{dt}(4t^{-1}) = 4(-1)t^{-2} = \frac{-4}{t^2} \)
Now, we calculate \( \frac{dy}{dx} \) using the chain rule:
\( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{-4}{t^2}}{4} \)
\( \implies \frac{dy}{dx} = \frac{-4}{t^2} \cdot \frac{1}{4} = \frac{-1}{t^2} \)
In simple words: We calculate the change rates of \( x \) and \( y \) with respect to \( t \). By dividing these rates, we find how \( y \) changes with \( x \), which comes out to be \( \frac{-1}{t^2} \).

Exam Tip: For expressions like \( \frac{4}{t} \), rewrite them as \( 4t^{-1} \) before differentiating to avoid errors in applying the power rule.

Question 5. If \( x = \cos \theta - \cos 2\theta, y = \sin \theta - \sin 2\theta \) then find \( \frac{dy}{dx} \).
Answer: We are given the equations:
\( x = \cos \theta - \cos 2\theta \)
\( y = \sin \theta - \sin 2\theta \)
First, we differentiate \( x \) and \( y \) with respect to \( \theta \):
\( \frac{dx}{d\theta} = \frac{d}{d\theta}(\cos \theta - \cos 2\theta) = -\sin \theta - (- \sin 2\theta) \cdot 2 = -\sin \theta + 2\sin 2\theta \)
\( \frac{dy}{d\theta} = \frac{d}{d\theta}(\sin \theta - \sin 2\theta) = \cos \theta - (\cos 2\theta) \cdot 2 = \cos \theta - 2\cos 2\theta \)
Now, we find \( \frac{dy}{dx} \) using the chain rule:
\( \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{\cos \theta - 2\cos 2\theta}{-\sin \theta + 2\sin 2\theta} \)
In simple words: We first determine how \( x \) and \( y \) change with respect to \( \theta \). Then, by dividing these rates, we can find how \( y \) changes with respect to \( x \).

Exam Tip: Be careful with the chain rule when differentiating \( \cos 2\theta \) and \( \sin 2\theta \). The derivative of \( \cos 2\theta \) is \( -\sin 2\theta \cdot 2 \) and for \( \sin 2\theta \) it is \( \cos 2\theta \cdot 2 \).

Question 6. If \( x = a(\theta - \sin \theta), y = a(1 + \cos \theta) \) then find \( \frac{dy}{dx} \).
Answer: We have the given equations:
\( x = a(\theta - \sin \theta) \)
\( y = a(1 + \cos \theta) \)
First, we differentiate \( x \) and \( y \) with respect to \( \theta \):
\( \frac{dx}{d\theta} = \frac{d}{d\theta}(a(\theta - \sin \theta)) = a(1 - \cos \theta) \)
\( \frac{dy}{d\theta} = \frac{d}{d\theta}(a(1 + \cos \theta)) = a(0 - \sin \theta) = -a \sin \theta \)
Now, we calculate \( \frac{dy}{dx} \) using the chain rule:
\( \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{-a \sin \theta}{a(1 - \cos \theta)} = \frac{-\sin \theta}{1 - \cos \theta} \)
Using half-angle identities: \( \sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \) and \( 1 - \cos \theta = 2 \sin^2 \frac{\theta}{2} \).
\( \implies \frac{dy}{dx} = \frac{-2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin^2 \frac{\theta}{2}} = \frac{-\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}} \)
\( \implies \frac{dy}{dx} = -\cot \frac{\theta}{2} \)
In simple words: First, we find how \( x \) and \( y \) change with \( \theta \). Then, we divide these rates and use trigonometric identities to simplify the result, which gives us \( -\cot \frac{\theta}{2} \).

Exam Tip: After finding the derivative in terms of \( \theta \), always check if trigonometric identities (like half-angle formulas) can simplify the expression further to a more compact form.

Question 7. If \( x = \frac{\sin^3 t}{\sqrt{\cos 2t}}, y = \frac{\cos^3 t}{\sqrt{\cos 2t}} \) then find \( \frac{dy}{dx} \).
Answer: We are given the equations:
\( x = \frac{\sin^3 t}{\sqrt{\cos 2t}} \) and \( y = \frac{\cos^3 t}{\sqrt{\cos 2t}} \)
We will use the quotient rule for differentiation, \( \frac{d}{dt}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \).

For \( x = \frac{\sin^3 t}{\sqrt{\cos 2t}} \):
Let \( u = \sin^3 t \) and \( v = \sqrt{\cos 2t} = (\cos 2t)^{1/2} \).
\( u' = \frac{d}{dt}(\sin^3 t) = 3\sin^2 t \cos t \)
\( v' = \frac{d}{dt}(\cos 2t)^{1/2} = \frac{1}{2}(\cos 2t)^{-1/2} (-\sin 2t)(2) = \frac{-\sin 2t}{\sqrt{\cos 2t}} \)
\( \frac{dx}{dt} = \frac{ (3\sin^2 t \cos t)(\sqrt{\cos 2t}) - (\sin^3 t)\left(\frac{-\sin 2t}{\sqrt{\cos 2t}}\right) }{ (\sqrt{\cos 2t})^2 } \)
\( \implies \frac{dx}{dt} = \frac{ 3\sin^2 t \cos t \sqrt{\cos 2t} + \frac{\sin^3 t \sin 2t}{\sqrt{\cos 2t}} }{ \cos 2t } \)
Multiply numerator and denominator by \( \sqrt{\cos 2t} \) to clear the fraction in the numerator:
\( \implies \frac{dx}{dt} = \frac{ 3\sin^2 t \cos t (\cos 2t) + \sin^3 t \sin 2t }{ (\cos 2t)^{3/2} } \)
Factor out \( \sin^2 t \):
\( \implies \frac{dx}{dt} = \frac{ \sin^2 t (3 \cos t \cos 2t + \sin t \sin 2t) }{ (\cos 2t)^{3/2} } \)

For \( y = \frac{\cos^3 t}{\sqrt{\cos 2t}} \):
Let \( u = \cos^3 t \) and \( v = \sqrt{\cos 2t} = (\cos 2t)^{1/2} \).
\( u' = \frac{d}{dt}(\cos^3 t) = 3\cos^2 t (-\sin t) = -3\sin t \cos^2 t \)
\( v' = \frac{d}{dt}(\cos 2t)^{1/2} = \frac{-\sin 2t}{\sqrt{\cos 2t}} \) (same as before)
\( \frac{dy}{dt} = \frac{ (-3\sin t \cos^2 t)(\sqrt{\cos 2t}) - (\cos^3 t)\left(\frac{-\sin 2t}{\sqrt{\cos 2t}}\right) }{ (\sqrt{\cos 2t})^2 } \)
\( \implies \frac{dy}{dt} = \frac{ -3\sin t \cos^2 t \sqrt{\cos 2t} + \frac{\cos^3 t \sin 2t}{\sqrt{\cos 2t}} }{ \cos 2t } \)
Multiply numerator and denominator by \( \sqrt{\cos 2t} \):
\( \implies \frac{dy}{dt} = \frac{ -3\sin t \cos^2 t (\cos 2t) + \cos^3 t \sin 2t }{ (\cos 2t)^{3/2} } \)
Factor out \( \cos^2 t \):
\( \implies \frac{dy}{dt} = \frac{ \cos^2 t (-3\sin t \cos 2t + \cos t \sin 2t) }{ (\cos 2t)^{3/2} } \)

Finally, we find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{ \frac{ \cos^2 t (-3\sin t \cos 2t + \cos t \sin 2t) }{ (\cos 2t)^{3/2} } }{ \frac{ \sin^2 t (3 \cos t \cos 2t + \sin t \sin 2t) }{ (\cos 2t)^{3/2} } } \)
The \( (\cos 2t)^{3/2} \) terms cancel out:
\( \implies \frac{dy}{dx} = \frac{ \cos^2 t (-3\sin t \cos 2t + \cos t \sin 2t) }{ \sin^2 t (3 \cos t \cos 2t + \sin t \sin 2t) } \)
Using \( \sin 2t = 2 \sin t \cos t \):
Numerator: \( \cos^2 t (-3\sin t \cos 2t + \cos t (2 \sin t \cos t)) = \cos^2 t \sin t (-3 \cos 2t + 2 \cos^2 t) \)
Denominator: \( \sin^2 t (3 \cos t \cos 2t + \sin t (2 \sin t \cos t)) = \sin^2 t \cos t (3 \cos 2t + 2 \sin^2 t) \)
\( \implies \frac{dy}{dx} = \frac{ \cos^2 t \sin t (-3 \cos 2t + 2 \cos^2 t) }{ \sin^2 t \cos t (3 \cos 2t + 2 \sin^2 t) } \)
\( \implies \frac{dy}{dx} = \frac{ \cos t (-3 \cos 2t + 2 \cos^2 t) }{ \sin t (3 \cos 2t + 2 \sin^2 t) } \)
This is a simplified form. Further simplification is possible using \( \cos 2t = 2 \cos^2 t - 1 \) or \( \cos 2t = 1 - 2 \sin^2 t \).
Numerator: \( \cos t (-3(2\cos^2 t - 1) + 2 \cos^2 t) = \cos t (-6\cos^2 t + 3 + 2 \cos^2 t) = \cos t (3 - 4 \cos^2 t) \)
Denominator: \( \sin t (3(1 - 2\sin^2 t) + 2 \sin^2 t) = \sin t (3 - 6\sin^2 t + 2 \sin^2 t) = \sin t (3 - 4 \sin^2 t) \)
We know \( \cos 3t = 4 \cos^3 t - 3 \cos t = \cos t (4 \cos^2 t - 3) \). So \( 3 - 4 \cos^2 t = -\frac{\cos 3t}{\cos t} \).
And \( \sin 3t = 3 \sin t - 4 \sin^3 t = \sin t (3 - 4 \sin^2 t) \). So \( 3 - 4 \sin^2 t = \frac{\sin 3t}{\sin t} \).
\( \implies \frac{dy}{dx} = \frac{ \cos t (-\frac{\cos 3t}{\cos t}) }{ \sin t (\frac{\sin 3t}{\sin t}) } = \frac{-\cos 3t}{\sin 3t} \)
\( \implies \frac{dy}{dx} = -\cot 3t \)
In simple words: We find the derivative of \( x \) and \( y \) with respect to \( t \) using the quotient rule, then divide them. After simplifying the expression using trigonometric identities, we get \( -\cot 3t \).

Exam Tip: For complex fractions involving trigonometric powers and square roots, apply the quotient rule methodically. Remember to simplify the expression using double and triple angle identities to reach the most concise form.

Question 8. If \( x = a[\cos t + \log \tan\frac{t}{2}], y = a \sin t \) then find \( \frac{dy}{dx} \).
Answer: We have the given parametric equations:
\( x = a\left[\cos t + \log \tan\frac{t}{2}\right] \)
\( y = a \sin t \)
First, we differentiate \( x \) and \( y \) with respect to \( t \):
\( \frac{dx}{dt} = a \frac{d}{dt}\left[\cos t + \log \tan\frac{t}{2}\right] \)
\( \implies \frac{dx}{dt} = a \left[-\sin t + \frac{1}{\tan\frac{t}{2}} \cdot \sec^2\frac{t}{2} \cdot \frac{1}{2}\right] \)
\( \implies \frac{dx}{dt} = a \left[-\sin t + \frac{\cos\frac{t}{2}}{\sin\frac{t}{2}} \cdot \frac{1}{\cos^2\frac{t}{2}} \cdot \frac{1}{2}\right] \)
\( \implies \frac{dx}{dt} = a \left[-\sin t + \frac{1}{2\sin\frac{t}{2}\cos\frac{t}{2}}\right] \)
We know that \( \sin t = 2\sin\frac{t}{2}\cos\frac{t}{2} \).
\( \implies \frac{dx}{dt} = a \left[-\sin t + \frac{1}{\sin t}\right] \)
\( \implies \frac{dx}{dt} = a \left[\frac{-\sin^2 t + 1}{\sin t}\right] \)
\( \implies \frac{dx}{dt} = a \frac{\cos^2 t}{\sin t} \)

Now for \( y \):
\( \frac{dy}{dt} = \frac{d}{dt}(a \sin t) = a \cos t \)

Finally, we find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{a \cos t}{a \frac{\cos^2 t}{\sin t}} \)
\( \implies \frac{dy}{dx} = \frac{\cos t \cdot \sin t}{\cos^2 t} = \frac{\sin t}{\cos t} \)
\( \implies \frac{dy}{dx} = \tan t \)
In simple words: We find the rate of change for \( x \) and \( y \) with respect to \( t \), simplifying \( \frac{dx}{dt} \) using trigonometric identities. Then, we divide \( \frac{dy}{dt} \) by \( \frac{dx}{dt} \) to get \( \tan t \).

Exam Tip: When dealing with logarithmic functions of trigonometric terms, remember the chain rule for derivatives of \( \log f(x) \) and \( \tan(\frac{x}{2}) \). Simplify expressions using fundamental trigonometric identities to reach the final answer.

Question 9. If \( x = a \sec \theta, y = b \tan \theta \) then find \( \frac{dy}{dx} \).
Answer: We have the given equations:
\( x = a \sec \theta \)
\( y = b \tan \theta \)
First, we differentiate \( x \) and \( y \) with respect to \( \theta \):
\( \frac{dx}{d\theta} = \frac{d}{d\theta}(a \sec \theta) = a \sec \theta \tan \theta \)
\( \frac{dy}{d\theta} = \frac{d}{d\theta}(b \tan \theta) = b \sec^2 \theta \)
Now, we find \( \frac{dy}{dx} \) using the chain rule:
\( \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{b \sec^2 \theta}{a \sec \theta \tan \theta} \)
\( \implies \frac{dy}{dx} = \frac{b}{a} \cdot \frac{\sec \theta}{\tan \theta} \)
We know that \( \sec \theta = \frac{1}{\cos \theta} \) and \( \tan \theta = \frac{\sin \theta}{\cos \theta} \).
\( \implies \frac{dy}{dx} = \frac{b}{a} \cdot \frac{\frac{1}{\cos \theta}}{\frac{\sin \theta}{\cos \theta}} = \frac{b}{a} \cdot \frac{1}{\sin \theta} \)
\( \implies \frac{dy}{dx} = \frac{b}{a} \operatorname{cosec} \theta \)
In simple words: We find how \( x \) and \( y \) change with \( \theta \). Then, we divide these rates and simplify the trigonometric terms to get \( \frac{b}{a} \operatorname{cosec} \theta \).

Exam Tip: Always simplify the final trigonometric expression using fundamental identities such as \( \sec \theta = \frac{1}{\cos \theta} \) and \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) to get the most compact form.

Question 10. If \( x = a(\cos \theta + \theta \sin \theta), y = a (\sin \theta - \theta \cos \theta) \) then find \( \frac{dy}{dx} \).
Answer: We are given the equations:
\( x = a(\cos \theta + \theta \sin \theta) \)
\( y = a (\sin \theta - \theta \cos \theta) \)
First, we differentiate \( x \) and \( y \) with respect to \( \theta \):
For \( x \):
\( \frac{dx}{d\theta} = a \frac{d}{d\theta}(\cos \theta + \theta \sin \theta) \)
\( \implies \frac{dx}{d\theta} = a (-\sin \theta + (1 \cdot \sin \theta + \theta \cdot \cos \theta)) \) (using product rule for \( \theta \sin \theta \))
\( \implies \frac{dx}{d\theta} = a (-\sin \theta + \sin \theta + \theta \cos \theta) \)
\( \implies \frac{dx}{d\theta} = a \theta \cos \theta \)
For \( y \):
\( \frac{dy}{d\theta} = a \frac{d}{d\theta}(\sin \theta - \theta \cos \theta) \)
\( \implies \frac{dy}{d\theta} = a (\cos \theta - (1 \cdot \cos \theta + \theta \cdot (-\sin \theta))) \) (using product rule for \( \theta \cos \theta \))
\( \implies \frac{dy}{d\theta} = a (\cos \theta - \cos \theta + \theta \sin \theta) \)
\( \implies \frac{dy}{d\theta} = a \theta \sin \theta \)
Now, we find \( \frac{dy}{dx} \) using the chain rule:
\( \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{a \theta \sin \theta}{a \theta \cos \theta} \)
\( \implies \frac{dy}{dx} = \frac{\sin \theta}{\cos \theta} \)
\( \implies \frac{dy}{dx} = \tan \theta \)
In simple words: We differentiate \( x \) and \( y \) separately with respect to \( \theta \), applying the product rule where needed. Then, we divide \( \frac{dy}{d\theta} \) by \( \frac{dx}{d\theta} \) to get \( \tan \theta \).

Exam Tip: When applying the product rule for terms like \( \theta \sin \theta \) or \( \theta \cos \theta \), be very careful with the signs and the differentiation of each factor. The cancellation of terms often simplifies these expressions significantly.

Question 11. If \( x = \sqrt{a^{\sin^{-1} t}} \) and \( y = \sqrt{a^{\cos^{-1} t}} \), show that \( \frac{dy}{dx} = \frac{-y}{x} \).
Answer: We have the given equations:
\( x = \sqrt{a^{\sin^{-1} t}} = (a^{\sin^{-1} t})^{1/2} = a^{\frac{1}{2} \sin^{-1} t} \)
\( y = \sqrt{a^{\cos^{-1} t}} = (a^{\cos^{-1} t})^{1/2} = a^{\frac{1}{2} \cos^{-1} t} \)
We will differentiate \( x \) and \( y \) with respect to \( t \). Recall that \( \frac{d}{dt}(a^u) = a^u \log a \frac{du}{dt} \).
For \( x \):
\( \frac{dx}{dt} = a^{\frac{1}{2} \sin^{-1} t} \cdot \log a \cdot \frac{d}{dt}\left(\frac{1}{2} \sin^{-1} t\right) \)
\( \implies \frac{dx}{dt} = a^{\frac{1}{2} \sin^{-1} t} \cdot \log a \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{1 - t^2}} \)
\( \implies \frac{dx}{dt} = \frac{x \log a}{2\sqrt{1 - t^2}} \)
For \( y \):
\( \frac{dy}{dt} = a^{\frac{1}{2} \cos^{-1} t} \cdot \log a \cdot \frac{d}{dt}\left(\frac{1}{2} \cos^{-1} t\right) \)
\( \implies \frac{dy}{dt} = a^{\frac{1}{2} \cos^{-1} t} \cdot \log a \cdot \frac{1}{2} \cdot \left(\frac{-1}{\sqrt{1 - t^2}}\right) \)
\( \implies \frac{dy}{dt} = \frac{-y \log a}{2\sqrt{1 - t^2}} \)
Now, we find \( \frac{dy}{dx} \) using the chain rule:
\( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{-y \log a}{2\sqrt{1 - t^2}}}{\frac{x \log a}{2\sqrt{1 - t^2}}} \)
The terms \( \log a \) and \( 2\sqrt{1 - t^2} \) cancel out:
\( \implies \frac{dy}{dx} = \frac{-y}{x} \)
Thus, we have shown that \( \frac{dy}{dx} = \frac{-y}{x} \).
In simple words: We first rewrite \( x \) and \( y \) using exponent rules. Then, we find the derivatives of \( x \) and \( y \) concerning \( t \). When we divide these derivatives, many terms cancel out, leaving us with \( \frac{-y}{x} \).

Exam Tip: For problems involving inverse trigonometric functions in exponents, remember the differentiation rules for \( \sin^{-1} t \) and \( \cos^{-1} t \), and the derivative of \( a^u \). Identifying \( x \) and \( y \) within the derivative expressions before cancellation can significantly simplify the process.