Samacheer Kalvi Class 11 Business Maths Solutions Chapter 5 Differential Calculus Exercise 5.8

Question 1. Find \( \frac{dy}{dx} \) of the following functions:
(i) \( x = ct, y = \frac{c}{t} \)
(ii) \( x = \log t, y = \sin t \)
(iii) \( x = a \cos^3\theta, y = a \sin^3\theta \)
(iv) \( x = a(\theta - \sin \theta), y = a(1 - \cos \theta) \)
Answer:
(i) Given functions are \( x = ct \) and \( y = \frac{c}{t} \).
First, we find the derivative of \( x \) with respect to \( t \):
\( \frac{dx}{dt} = \frac{d}{dt}(ct) = c \)
Next, we find the derivative of \( y \) with respect to \( t \):
\( y = ct^{-1} \)
\( \frac{dy}{dt} = \frac{d}{dt}(ct^{-1}) = c(-1)t^{-2} = -\frac{c}{t^2} \)
Now, we find \( \frac{dy}{dx} \) using the chain rule:
\( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \)
\( \frac{dy}{dx} = \frac{-\frac{c}{t^2}}{c} \)
\( \implies \frac{dy}{dx} = -\frac{1}{t^2} \)

(ii) Given functions are \( x = \log t \) and \( y = \sin t \).
First, we find the derivative of \( x \) with respect to \( t \):
\( \frac{dx}{dt} = \frac{d}{dt}(\log t) = \frac{1}{t} \)
Next, we find the derivative of \( y \) with respect to \( t \):
\( \frac{dy}{dt} = \frac{d}{dt}(\sin t) = \cos t \)
Now, we find \( \frac{dy}{dx} \) using the chain rule:
\( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \)
\( \frac{dy}{dx} = \frac{\cos t}{\frac{1}{t}} \)
\( \implies \frac{dy}{dx} = t \cos t \)

(iii) Given functions are \( x = a \cos^3\theta \) and \( y = a \sin^3\theta \).
First, we find the derivative of \( x \) with respect to \( \theta \):
\( \frac{dx}{d\theta} = \frac{d}{d\theta}(a \cos^3\theta) = a \cdot 3 \cos^2\theta (-\sin\theta) = -3a \cos^2\theta \sin\theta \)
Next, we find the derivative of \( y \) with respect to \( \theta \):
\( \frac{dy}{d\theta} = \frac{d}{d\theta}(a \sin^3\theta) = a \cdot 3 \sin^2\theta (\cos\theta) = 3a \sin^2\theta \cos\theta \)
Now, we find \( \frac{dy}{dx} \) using the chain rule:
\( \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \)
\( \frac{dy}{dx} = \frac{3a \sin^2\theta \cos\theta}{-3a \cos^2\theta \sin\theta} \)
We can simplify the expression by cancelling common terms \( 3a \sin\theta \cos\theta \):
\( \implies \frac{dy}{dx} = \frac{\sin\theta}{-\cos\theta} \)
\( \implies \frac{dy}{dx} = -\tan\theta \)

(iv) Given functions are \( x = a(\theta - \sin \theta) \) and \( y = a(1 - \cos \theta) \).
First, we find the derivative of \( x \) with respect to \( \theta \):
\( \frac{dx}{d\theta} = \frac{d}{d\theta}(a(\theta - \sin \theta)) = a(1 - \cos \theta) \)
Next, we find the derivative of \( y \) with respect to \( \theta \):
\( \frac{dy}{d\theta} = \frac{d}{d\theta}(a(1 - \cos \theta)) = a(0 - (-\sin \theta)) = a \sin \theta \)
Now, we find \( \frac{dy}{dx} \) using the chain rule:
\( \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \)
\( \frac{dy}{dx} = \frac{a \sin \theta}{a(1 - \cos \theta)} \)
We can use trigonometric half-angle identities to simplify. We know that \( \sin \theta = 2 \sin\frac{\theta}{2} \cos\frac{\theta}{2} \) and \( 1 - \cos \theta = 2 \sin^2\frac{\theta}{2} \).
\( \implies \frac{dy}{dx} = \frac{2 \sin\frac{\theta}{2} \cos\frac{\theta}{2}}{2 \sin^2\frac{\theta}{2}} \)
\( \implies \frac{dy}{dx} = \frac{\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}} \)
\( \implies \frac{dy}{dx} = \cot\frac{\theta}{2} \)
In simple words: To find \( \frac{dy}{dx} \) for functions given in terms of another variable (like \( t \) or \( \theta \)), we first find the derivatives of \( x \) and \( y \) with respect to that variable. Then, we divide \( \frac{dy}{dt} \) by \( \frac{dx}{dt} \) (or \( \frac{dy}{d\theta} \) by \( \frac{dx}{d\theta} \)). Sometimes, using simple trigonometric identities helps make the answer shorter.

🎯 Exam Tip: When dealing with parametric differentiation, clearly state the chain rule formula \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \) or \( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \) and correctly apply differentiation rules to each part. Pay attention to simplification using trigonometric identities if necessary.

Question 2. Differentiate \( \sin^3x \) with respect to \( \cos^3x \).
Answer: To differentiate \( \sin^3x \) with respect to \( \cos^3x \), we treat each as a separate function of \( x \).
Let \( u = \sin^3x \) and \( v = \cos^3x \). We need to find \( \frac{du}{dv} \).
First, find the derivative of \( u \) with respect to \( x \):
\( u = (\sin x)^3 \)
\( \frac{du}{dx} = 3(\sin x)^2 \cdot \frac{d}{dx}(\sin x) = 3 \sin^2 x \cos x \)
Next, find the derivative of \( v \) with respect to \( x \):
\( v = (\cos x)^3 \)
\( \frac{dv}{dx} = 3(\cos x)^2 \cdot \frac{d}{dx}(\cos x) = 3 \cos^2 x (-\sin x) = -3 \cos^2 x \sin x \)
Now, use the chain rule to find \( \frac{du}{dv} \):
\( \frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}} \)
\( \frac{du}{dv} = \frac{3 \sin^2 x \cos x}{-3 \cos^2 x \sin x} \)
We can simplify the expression by cancelling common terms \( 3 \sin x \cos x \):
\( \implies \frac{du}{dv} = \frac{\sin x}{-\cos x} \)
\( \implies \frac{du}{dv} = -\tan x \)
In simple words: When asked to differentiate one function with respect to another, we find the derivative of each function separately with respect to \( x \). Then, we divide the first derivative by the second. This helps us find how one function changes as the other function changes.

🎯 Exam Tip: Remember to apply the chain rule correctly for composite functions like \( (\sin x)^3 \). After finding \( \frac{du}{dx} \) and \( \frac{dv}{dx} \), simplify the fraction \( \frac{du}{dv} \) by cancelling out common terms carefully to get the final answer.

Question 3. Differentiate \( \sin^2x \) with respect to \( x^2 \).
Answer: To differentiate \( \sin^2x \) with respect to \( x^2 \), we treat each as a separate function of \( x \).
Let \( u = \sin^2x \) and \( v = x^2 \). We need to find \( \frac{du}{dv} \).
First, find the derivative of \( u \) with respect to \( x \):
\( u = (\sin x)^2 \)
\( \frac{du}{dx} = 2(\sin x) \cdot \frac{d}{dx}(\sin x) = 2 \sin x \cos x \)
Using the double angle identity, \( 2 \sin x \cos x = \sin 2x \):
\( \implies \frac{du}{dx} = \sin 2x \)
Next, find the derivative of \( v \) with respect to \( x \):
\( \frac{dv}{dx} = \frac{d}{dx}(x^2) = 2x \)
Now, use the chain rule to find \( \frac{du}{dv} \):
\( \frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}} \)
\( \frac{du}{dv} = \frac{\sin 2x}{2x} \)
In simple words: We want to know how \( \sin^2x \) changes when \( x^2 \) changes. We do this by first finding how each function changes with \( x \). Then, we divide the derivative of \( \sin^2x \) by the derivative of \( x^2 \).

🎯 Exam Tip: For differentiation problems, always look for opportunities to simplify using trigonometric identities like \( 2 \sin x \cos x = \sin 2x \). This helps present the answer in its most standard form.