Differentiation: Top 10 Important Board Exam Sums
Differentiation is a high-weightage chapter in the 12th Mathematics syllabus. Below are 10 highly probable questions covering Chain Rule, Inverse Trigonometric Functions, Logarithmic Differentiation, Implicit Functions, and Parametric Forms.
Q1. Differentiate the following w.r.t. \(x\): \( y = \sin(\log x) \)
Solution:
Given function: $$ y = \sin(\log x) $$
Differentiating both sides with respect to \(x\):
$$ \frac{dy}{dx} = \frac{d}{dx}[\sin(\log x)] $$
Using Chain Rule: First differentiate sine, then differentiate the inner function \((\log x)\).
$$ \frac{dy}{dx} = \cos(\log x) \cdot \frac{d}{dx}(\log x) $$
We know that \( \frac{d}{dx}(\log x) = \frac{1}{x} \).
$$ \frac{dy}{dx} = \cos(\log x) \cdot \frac{1}{x} $$
$$ \therefore \frac{dy}{dx} = \frac{\cos(\log x)}{x} $$
Q2. Differentiate \( y = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \) w.r.t. \(x\).
Solution:
To simplify this Inverse Trigonometric function, we use substitution.
Put \( x = \tan\theta \), which implies \( \theta = \tan^{-1}x \).
Put \( x = \tan\theta \), which implies \( \theta = \tan^{-1}x \).
Substitute in the given equation:
$$ y = \tan^{-1}\left(\frac{2\tan\theta}{1-\tan^2\theta}\right) $$
Using the formula \( \tan 2\theta = \frac{2\tan\theta}{1-\tan^2\theta} \):
$$ y = \tan^{-1}(\tan 2\theta) $$
$$ y = 2\theta $$
Resubstitute the value of \(\theta\):
$$ y = 2\tan^{-1}x $$
Differentiating w.r.t \(x\):
$$ \frac{dy}{dx} = 2 \cdot \frac{d}{dx}(\tan^{-1}x) $$
$$ \frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2} $$
$$ \therefore \frac{dy}{dx} = \frac{2}{1+x^2} $$
Q3. If \( y = x^x \), find \( \frac{dy}{dx} \).
Solution:
This is a function raised to the power of a function. We must use Logarithmic Differentiation.
$$ y = x^x $$
Taking log on both sides:
$$ \log y = \log(x^x) $$
Using property \( \log(a^b) = b \log a \):
$$ \log y = x \cdot \log x $$
Differentiating w.r.t \(x\) using Product Rule on RHS:
$$ \frac{1}{y} \frac{dy}{dx} = x \cdot \frac{d}{dx}(\log x) + \log x \cdot \frac{d}{dx}(x) $$
$$ \frac{1}{y} \frac{dy}{dx} = x \cdot \frac{1}{x} + \log x \cdot (1) $$
$$ \frac{1}{y} \frac{dy}{dx} = 1 + \log x $$
Move \(y\) to the RHS:
$$ \frac{dy}{dx} = y(1 + \log x) $$
$$ \therefore \frac{dy}{dx} = x^x(1 + \log x) $$
Q4. Find \( \frac{dy}{dx} \) if \( x^2 + y^2 = \log(xy) \).
Solution:
Simplify the RHS using log properties:
$$ x^2 + y^2 = \log x + \log y $$
Differentiate w.r.t \(x\) (Implicit Differentiation):
$$ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(\log x) + \frac{d}{dx}(\log y) $$
$$ 2x + 2y\frac{dy}{dx} = \frac{1}{x} + \frac{1}{y}\frac{dy}{dx} $$
Group terms containing \( \frac{dy}{dx} \) on one side:
$$ 2y\frac{dy}{dx} - \frac{1}{y}\frac{dy}{dx} = \frac{1}{x} - 2x $$
Factor out \( \frac{dy}{dx} \):
$$ \frac{dy}{dx}\left( 2y - \frac{1}{y} \right) = \frac{1 - 2x^2}{x} $$
$$ \frac{dy}{dx}\left( \frac{2y^2 - 1}{y} \right) = \frac{1 - 2x^2}{x} $$
Isolate \( \frac{dy}{dx} \):
$$ \frac{dy}{dx} = \frac{1 - 2x^2}{x} \cdot \frac{y}{2y^2 - 1} $$
$$ \therefore \frac{dy}{dx} = \frac{y(1 - 2x^2)}{x(2y^2 - 1)} $$
Q5. Find \( \frac{dy}{dx} \) if \( x = a(\theta - \sin\theta) \) and \( y = a(1 - \cos\theta) \).
Solution:
This is a Parametric Function.
First, differentiate \(x\) w.r.t \(\theta\):
$$ \frac{dx}{d\theta} = a(1 - \cos\theta) $$
Now, differentiate \(y\) w.r.t \(\theta\):
$$ \frac{dy}{d\theta} = a(0 - (-\sin\theta)) = a\sin\theta $$
By parametric rule:
$$ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} $$
$$ \frac{dy}{dx} = \frac{a\sin\theta}{a(1 - \cos\theta)} = \frac{\sin\theta}{1 - \cos\theta} $$
Simplify using Half-Angle formulas:
\( \sin\theta = 2\sin(\theta/2)\cos(\theta/2) \)
\( 1 - \cos\theta = 2\sin^2(\theta/2) \)
\( \sin\theta = 2\sin(\theta/2)\cos(\theta/2) \)
\( 1 - \cos\theta = 2\sin^2(\theta/2) \)
$$ \frac{dy}{dx} = \frac{2\sin(\theta/2)\cos(\theta/2)}{2\sin^2(\theta/2)} $$
$$ \frac{dy}{dx} = \frac{\cos(\theta/2)}{\sin(\theta/2)} $$
$$ \therefore \frac{dy}{dx} = \cot\left(\frac{\theta}{2}\right) $$
Q6. 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 \( y = e^{m \tan^{-1} x} \).
Differentiate w.r.t \(x\): $$ \frac{dy}{dx} = e^{m \tan^{-1} x} \cdot \frac{d}{dx}(m \tan^{-1} x) $$ $$ \frac{dy}{dx} = y \cdot m \cdot \frac{1}{1+x^2} $$
Differentiate w.r.t \(x\): $$ \frac{dy}{dx} = e^{m \tan^{-1} x} \cdot \frac{d}{dx}(m \tan^{-1} x) $$ $$ \frac{dy}{dx} = y \cdot m \cdot \frac{1}{1+x^2} $$
Rearrange terms to avoid quotient rule:
$$ (1+x^2)\frac{dy}{dx} = my $$
Differentiate again w.r.t \(x\) (Using Product Rule on LHS):
$$ (1+x^2) \cdot \frac{d}{dx}\left(\frac{dy}{dx}\right) + \frac{dy}{dx} \cdot \frac{d}{dx}(1+x^2) = m \cdot \frac{dy}{dx} $$
$$ (1+x^2)\frac{d^2y}{dx^2} + \frac{dy}{dx}(2x) = m\frac{dy}{dx} $$
Bring all terms to LHS:
$$ (1+x^2)\frac{d^2y}{dx^2} + 2x\frac{dy}{dx} - m\frac{dy}{dx} = 0 $$
$$ (1+x^2)\frac{d^2y}{dx^2} + (2x-m)\frac{dy}{dx} = 0 $$
Hence Proved.
Hence Proved.
Q7. If \( y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \dots \infty}}} \), find \( \frac{dy}{dx} \).
Solution:
Since the series is infinite, the term inside the first square root is effectively equal to \(y\) itself.
$$ y = \sqrt{\sin x + y} $$
Squaring both sides to remove the root:
$$ y^2 = \sin x + y $$
Differentiate w.r.t \(x\) (Implicit function):
$$ 2y\frac{dy}{dx} = \cos x + \frac{dy}{dx} $$
Collect \( \frac{dy}{dx} \) terms:
$$ 2y\frac{dy}{dx} - \frac{dy}{dx} = \cos x $$
$$ \frac{dy}{dx}(2y - 1) = \cos x $$
$$ \therefore \frac{dy}{dx} = \frac{\cos x}{2y - 1} $$
Q8. Differentiate \( \cos^{-1}(4x^3 - 3x) \) w.r.t \(x\).
Solution:
Let \( y = \cos^{-1}(4x^3 - 3x) \).
Use substitution: Put \( x = \cos\theta \), so \( \theta = \cos^{-1}x \).
Use substitution: Put \( x = \cos\theta \), so \( \theta = \cos^{-1}x \).
Substitute in the equation:
$$ y = \cos^{-1}(4\cos^3\theta - 3\cos\theta) $$
Using the Triple Angle formula \( \cos 3\theta = 4\cos^3\theta - 3\cos\theta \):
$$ y = \cos^{-1}(\cos 3\theta) $$
$$ y = 3\theta $$
Resubstitute \( \theta \):
$$ y = 3\cos^{-1}x $$
Differentiate w.r.t \(x\):
$$ \frac{dy}{dx} = 3 \cdot \frac{-1}{\sqrt{1-x^2}} $$
$$ \therefore \frac{dy}{dx} = \frac{-3}{\sqrt{1-x^2}} $$
Q9. Differentiate \( e^x \) w.r.t \( \sin x \).
Solution:
Let \( u = e^x \) and \( v = \sin x \).
We need to find \( \frac{du}{dv} \).
We need to find \( \frac{du}{dv} \).
First find \( \frac{du}{dx} \):
$$ \frac{du}{dx} = \frac{d}{dx}(e^x) = e^x $$
Next find \( \frac{dv}{dx} \):
$$ \frac{dv}{dx} = \frac{d}{dx}(\sin x) = \cos x $$
Now, \( \frac{du}{dv} = \frac{du/dx}{dv/dx} \):
$$ \frac{du}{dv} = \frac{e^x}{\cos x} $$
$$ \therefore \text{Derivative} = e^x \sec x $$
Q10. If \( x^y = e^{x-y} \), show that \( \frac{dy}{dx} = \frac{\log x}{(1+\log x)^2} \).
Solution:
Given \( x^y = e^{x-y} \).
Taking log on both sides: $$ y \log x = (x-y) \log e $$
Taking log on both sides: $$ y \log x = (x-y) \log e $$
Since \( \log e = 1 \):
$$ y \log x = x - y $$
$$ y \log x + y = x $$
$$ y(1 + \log x) = x $$
Express \(y\) purely in terms of \(x\):
$$ y = \frac{x}{1 + \log x} $$
Differentiate using Quotient Rule \( \left( \frac{u}{v} \right)' = \frac{v u' - u v'}{v^2} \):
$$ \frac{dy}{dx} = \frac{(1+\log x)\cdot(1) - x\cdot(\frac{1}{x})}{(1+\log x)^2} $$
Simplify the numerator:
$$ \frac{dy}{dx} = \frac{1 + \log x - 1}{(1+\log x)^2} $$
$$ \frac{dy}{dx} = \frac{\log x}{(1+\log x)^2} $$
Hence Proved.
Hence Proved.
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