The following list contains 10 highly probable derivative problems covering key concepts like Chain Rule, Implicit Differentiation, Logarithmic Differentiation, Parametric Functions, and Second Order Derivatives suitable for the HSC Board Examination.
Q1. Differentiate the following w.r.t. \(x\): \( y = \tan^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right) \)
Solution:
We use the half-angle trigonometric formulas:
\( 1 - \cos x = 2\sin^2(x/2) \) and \( 1 + \cos x = 2\cos^2(x/2) \).
Substitute these into the given function:
$$ y = \tan^{-1}\left(\sqrt{\frac{2\sin^2(x/2)}{2\cos^2(x/2)}}\right) $$
$$ y = \tan^{-1}\left(\sqrt{\tan^2(x/2)}\right) $$
$$ y = \tan^{-1}(\tan(x/2)) $$
Since \( \tan^{-1}(\tan \theta) = \theta \), we get:
\( y = \frac{x}{2} \)
Differentiating w.r.t \(x\):
$$ \frac{dy}{dx} = \frac{1}{2} $$
Q2. If \( x^m y^n = (x+y)^{m+n} \), prove that \( \frac{dy}{dx} = \frac{y}{x} \).
Solution:
Taking log on both sides:
\( \log(x^m y^n) = \log(x+y)^{m+n} \)
\( m \log x + n \log y = (m+n) \log(x+y) \)
Differentiating both sides w.r.t \(x\):
$$ \frac{m}{x} + \frac{n}{y}\frac{dy}{dx} = \frac{m+n}{x+y} \left(1 + \frac{dy}{dx}\right) $$
Group terms containing \( \frac{dy}{dx} \):
$$ \frac{n}{y}\frac{dy}{dx} - \frac{m+n}{x+y}\frac{dy}{dx} = \frac{m+n}{x+y} - \frac{m}{x} $$
$$ \frac{dy}{dx} \left[ \frac{n(x+y) - y(m+n)}{y(x+y)} \right] = \left[ \frac{x(m+n) - m(x+y)}{x(x+y)} \right] $$
$$ \frac{dy}{dx} \left[ \frac{nx + ny - my - ny}{y} \right] = \left[ \frac{mx + nx - mx - my}{x} \right] $$
$$ \frac{dy}{dx} \left[ \frac{nx - my}{y} \right] = \left[ \frac{nx - my}{x} \right] $$
Canceling \( (nx - my) \) from both sides:
$$ \frac{dy}{dx} = \frac{y}{x} $$ (Hence Proved)
Q3. Find \( \frac{dy}{dx} \) if \( y = x^x + (\sin x)^x \).
Solution:
Let \( y = u + v \), where \( u = x^x \) and \( v = (\sin x)^x \).
Then \( \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \).
Part 1 (Solving u): \( u = x^x \). Taking log: \( \log u = x \log x \).
Diff w.r.t \(x\): \( \frac{1}{u}\frac{du}{dx} = x(\frac{1}{x}) + \log x(1) \).
\( \frac{du}{dx} = x^x(1 + \log x) \).
Part 2 (Solving v): \( v = (\sin x)^x \). Taking log: \( \log v = x \log(\sin x) \).
Diff w.r.t \(x\): \( \frac{1}{v}\frac{dv}{dx} = x \cdot \frac{1}{\sin x} \cdot \cos x + \log(\sin x) \cdot 1 \).
\( \frac{dv}{dx} = (\sin x)^x [x \cot x + \log(\sin x)] \).
Combine both parts:
$$ \frac{dy}{dx} = x^x(1 + \log x) + (\sin x)^x [x \cot x + \log(\sin x)] $$
Q4. If \( x = a(\theta + \sin\theta) \) and \( y = a(1 - \cos\theta) \), find \( \frac{dy}{dx} \) at \( \theta = \frac{\pi}{2} \).
Solution:
Differentiate \(x\) w.r.t \(\theta\):
\( \frac{dx}{d\theta} = a(1 + \cos\theta) \).
Differentiate \(y\) w.r.t \(\theta\):
\( \frac{dy}{d\theta} = a(0 - (-\sin\theta)) = a \sin\theta \).
Now, \( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \):
$$ \frac{dy}{dx} = \frac{a \sin\theta}{a(1 + \cos\theta)} = \frac{\sin\theta}{1 + \cos\theta} $$
Using half-angle formulas: \( \frac{2\sin(\theta/2)\cos(\theta/2)}{2\cos^2(\theta/2)} = \tan(\theta/2) \).
At \( \theta = \frac{\pi}{2} \):
$$ \left(\frac{dy}{dx}\right)_{\theta=\pi/2} = \tan\left(\frac{\pi}{4}\right) = 1 $$
Q5. Differentiate \( \sin^{-1}\left(\frac{2x}{1+x^2}\right) \) with respect to \( \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \).
Solution:
Let \( u = \sin^{-1}(\frac{2x}{1+x^2}) \) and \( v = \cos^{-1}(\frac{1-x^2}{1+x^2}) \).
Substitute \( x = \tan\theta \), then \( \theta = \tan^{-1}x \).
For \( u \): \( \sin^{-1}(\sin 2\theta) = 2\theta = 2\tan^{-1}x \).
For \( v \): \( \cos^{-1}(\cos 2\theta) = 2\theta = 2\tan^{-1}x \).
We need to find \( \frac{du}{dv} \).
$$ \frac{du}{dv} = \frac{du/dx}{dv/dx} $$
$$ \frac{du}{dx} = \frac{2}{1+x^2} $$
$$ \frac{dv}{dx} = \frac{2}{1+x^2} $$
$$ \frac{du}{dv} = \frac{\frac{2}{1+x^2}}{\frac{2}{1+x^2}} = 1 $$
Q6. If \( y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \dots \infty}}} \), find \( \frac{dy}{dx} \).
Solution:
The given expression can be written as:
\( y = \sqrt{\sin x + y} \)
Squaring both sides:
\( y^2 = \sin x + y \)
Differentiating w.r.t \(x\):
\( 2y \frac{dy}{dx} = \cos x + \frac{dy}{dx} \)
\( \frac{dy}{dx}(2y - 1) = \cos x \)
$$ \frac{dy}{dx} = \frac{\cos x}{2y - 1} $$
Q7. If \( y = e^{m \tan^{-1}x} \), show that \( (1+x^2)\frac{d^2y}{dx^2} + (2x-m)\frac{dy}{dx} = 0 \).
Solution:
Given \( y = e^{m \tan^{-1}x} \).
Differentiating w.r.t \(x\):
\( \frac{dy}{dx} = e^{m \tan^{-1}x} \cdot \frac{m}{1+x^2} \)
\( (1+x^2)\frac{dy}{dx} = my \) (Since \( e^{m \tan^{-1}x} = y \))
Differentiating again w.r.t \(x\) using Product Rule:
\( (1+x^2)\frac{d^2y}{dx^2} + \frac{dy}{dx}(2x) = m \frac{dy}{dx} \)
Rearranging terms:
\( (1+x^2)\frac{d^2y}{dx^2} + 2x\frac{dy}{dx} - m\frac{dy}{dx} = 0 \)
$$ (1+x^2)\frac{d^2y}{dx^2} + (2x-m)\frac{dy}{dx} = 0 $$ (Hence Proved)
Q8. If \( x = a \cos t \), \( y = b \sin t \), find \( \frac{d^2y}{dx^2} \).
Solution:
1. Find \( \frac{dx}{dt} = -a \sin t \) and \( \frac{dy}{dt} = b \cos t \).
2. Find \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{b \cos t}{-a \sin t} = -\frac{b}{a} \cot t \).
3. Differentiate \( \frac{dy}{dx} \) w.r.t \( x \):
\( \frac{d^2y}{dx^2} = \frac{d}{dx}(-\frac{b}{a} \cot t) \)
Use Chain Rule: \( \frac{d}{dt}(-\frac{b}{a} \cot t) \cdot \frac{dt}{dx} \)
\( = -\frac{b}{a}(-\text{cosec}^2 t) \cdot \frac{1}{dx/dt} \)
\( = \frac{b}{a} \text{cosec}^2 t \cdot \frac{1}{-a \sin t} \)
$$ \frac{d^2y}{dx^2} = -\frac{b}{a^2} \text{cosec}^3 t $$
Q9. Differentiate \( (\log x)^x \) with respect to \(x\).
Solution:
Let \( y = (\log x)^x \). Take log on both sides:
\( \log y = x \cdot \log(\log x) \)
Differentiating w.r.t \(x\) using Product Rule:
\( \frac{1}{y} \frac{dy}{dx} = x \cdot \frac{d}{dx}(\log(\log x)) + \log(\log x) \cdot \frac{d}{dx}(x) \)
\( \frac{1}{y} \frac{dy}{dx} = x \cdot \frac{1}{\log x} \cdot \frac{1}{x} + \log(\log x) \cdot 1 \)
\( \frac{1}{y} \frac{dy}{dx} = \frac{1}{\log x} + \log(\log x) \)
Multiply by \( y \):
$$ \frac{dy}{dx} = (\log x)^x \left[ \frac{1}{\log x} + \log(\log x) \right] $$
Q10. Differentiate \( e^{x} + e^y = e^{x+y} \) with respect to \(x\).
Solution:
Given \( e^x + e^y = e^{x+y} \).
Differentiating w.r.t \(x\):
\( e^x + e^y \frac{dy}{dx} = e^{x+y} (1 + \frac{dy}{dx}) \)
Substitute \( e^{x+y} = e^x + e^y \) into the RHS:
\( e^x + e^y \frac{dy}{dx} = (e^x + e^y)(1 + \frac{dy}{dx}) \)
\( e^x + e^y \frac{dy}{dx} = e^x + e^x \frac{dy}{dx} + e^y + e^y \frac{dy}{dx} \)
Simplify: \( e^x + e^y \frac{dy}{dx} = e^x + e^y + \frac{dy}{dx}(e^x + e^y) \)
Wait, let's solve directly from step 1 without substitution for easier grouping:
\( e^x + e^y \frac{dy}{dx} = e^{x+y} + e^{x+y} \frac{dy}{dx} \)
\( \frac{dy}{dx}(e^y - e^{x+y}) = e^{x+y} - e^x \)
\( \frac{dy}{dx} = \frac{e^{x+y} - e^x}{e^y - e^{x+y}} \)
Now use \( e^{x+y} = e^x + e^y \):
\( \text{Numerator: } (e^x + e^y) - e^x = e^y \)
\( \text{Denominator: } e^y - (e^x + e^y) = -e^x \)
$$ \frac{dy}{dx} = -\frac{e^y}{e^x} = -e^{y-x} $$