JEE MATHS PRACTICE - MAINS TO ADVANCED - CALCULUS - DIFFERENTIATION AND APPLICATION OF DERIVATIVES

I will now provide 25 concept-based solved examples on Differentiation and Application of Derivatives, arranged from easy to challenging levels.

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PART 1: Basic Problems (Easy Level)

1. Find $\frac{d}{dx} (x^3 - 4x^2 + 5x - 7)$.

Solution:

$$\frac{d}{dx} (x^3 - 4x^2 + 5x - 7) = 3x^2 - 8x + 5.$$

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2. Differentiate $y = e^x + \ln x$.

Solution:

$$\frac{dy}{dx} = e^x + \frac{1}{x}.$$

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3. Differentiate $y = \sin x + \cos x$.

Solution:

$$\frac{dy}{dx} = \cos x - \sin x.$$

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4. Differentiate $y = x^x$.

Solution:
Taking logarithm,

$$\ln y = x \ln x.$$

Differentiating,

$$\frac{1}{y} \frac{dy}{dx} = \ln x + 1.$$ $$\frac{dy}{dx} = x^x (\ln x + 1).$$

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5. Differentiate $y = \frac{1}{\sqrt{x}}$.

Solution:
Rewriting,

$$y = x^{-1/2}.$$ $$\frac{dy}{dx} = -\frac{1}{2} x^{-3/2} = -\frac{1}{2\sqrt{x^3}}.$$

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PART 2: Intermediate Level Problems

6. Find $f'(x)$ for $f(x) = x^3 e^x$.

Solution:
Using product rule,

$$f'(x) = 3x^2 e^x + x^3 e^x.$$

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7. Differentiate $y = \tan^{-1} (x^2)$.

Solution:

$$\frac{dy}{dx} = \frac{1}{1 + x^4} \cdot 2x = \frac{2x}{1 + x^4}.$$

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8. Differentiate y = x^x^x.

Solution:
Taking log,

$$\ln y = x^x \ln x.$$ $$\frac{dy}{dx} = y \left[ x^x (\ln x + 1) + x^{x-1} \right].$$

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9. Find the derivative of $y = x^3 \sin x$.

Solution:
Using product rule,

$$\frac{dy}{dx} = 3x^2 \sin x + x^3 \cos x.$$

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10. Differentiate $f(x) = e^{\sin x}$.

Solution:
Using chain rule,

$$\frac{df}{dx} = e^{\sin x} \cos x.$$

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PART 3: Application of Derivatives (Advanced Level Problems)

11. Find the slope of the normal to the curve $y = x^3 - 4x$ at $x = 2$.

Solution:

$$\frac{dy}{dx} = 3x^2 - 4.$$

At $x = 2$,

$$\frac{dy}{dx} = 3(4) - 4 = 8.$$

Normal slope = $-\frac{1}{8}$.

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12. Find the equation of the tangent to $y = e^x + x^2$ at $x = 0$.

Solution:

$$\frac{dy}{dx} = e^x + 2x.$$

At $x = 0$, $y = 1$, slope = $1$.
Equation:

$$y - 1 = (x - 0) \Rightarrow y = x + 1.$$

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13. Find the critical points of $f(x) = x^4 - 4x^3$.

Solution:

$$f'(x) = 4x^3 - 12x^2.$$

Setting $f'(x) = 0$,

$$4x^2 (x - 3) = 0 \Rightarrow x = 0, 3.$$

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14. Find the intervals of increase/decrease for $f(x) = x^3 - 3x^2 + 4$.

Solution:

$$f'(x) = 3x^2 - 6x.$$

Setting $f'(x) = 0$,

$$3x(x-2) = 0 \Rightarrow x = 0, 2.$$

Sign test:

• $f'(x) > 0$ in $(-\infty,0) \cup (2,\infty)$ (increasing).
• $f'(x) < 0$ in $(0,2)$ (decreasing).

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15. Find the local maxima and minima for $f(x) = x^3 - 6x^2 + 9x + 1$.

Solution:

$$f'(x) = 3x^2 - 12x + 9.$$

Setting $f'(x) = 0$,

$$3(x-1)(x-3) = 0 \Rightarrow x = 1, 3.$$

Using second derivative test,

$$f''(x) = 6x - 12.$$

At $x = 1$, $f''(1) = -6$ (maxima).
At $x = 3$, $f''(3) = 6$ (minima).

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PART 4: Challenging Problems

16-25: Higher-order problems involving curve sketching, optimization, and L'Hôpital’s Rule

These will involve concavity, point of inflection, Newton’s method, second derivative test, and real-world applications like maximizing area, minimizing cost, and velocity problems.

Here are 10 challenging problems (16-25) on Differentiation and Application of Derivatives, covering curve sketching, optimization, concavity, Newton’s method, and real-world applications.

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PART 4: Challenging Problems (Advanced Level)

16. Find the points of inflection for $f(x) = x^4 - 4x^3 + 6x^2 - 4x$.

Solution:
Inflection points occur where $f''(x) = 0$.
First derivative:

$$f'(x) = 4x^3 - 12x^2 + 12x - 4.$$

Second derivative:

$$f''(x) = 12x^2 - 24x + 12.$$

Setting $f''(x) = 0$,

$$12(x^2 - 2x + 1) = 0.$$ $$12(x-1)^2 = 0 \Rightarrow x = 1.$$

Since second derivative changes sign around $x = 1$, the point of inflection is $x = 1$.

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17. Find the absolute maximum and minimum of $f(x) = x^3 - 6x^2 + 9x + 2$ in $[0,4]$.

Solution:
First derivative:

$$f'(x) = 3x^2 - 12x + 9.$$

Setting $f'(x) = 0$,

$$3(x-1)(x-3) = 0 \Rightarrow x = 1, 3.$$

Checking function values:

$$f(0) = 2, \quad f(1) = 6, \quad f(3) = 2, \quad f(4) = 6.$$

Absolute maximum: $6$ at $x = 1, 4$.
Absolute minimum: $2$ at $x = 0, 3$.

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18. Use Newton’s method to approximate the root of $f(x) = x^3 - 2x - 5$ starting at $x_0 = 2$.

Solution:
Newton’s formula:

$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}.$$

First derivative:

$$f'(x) = 3x^2 - 2.$$

Iteration 1:

$$x_1 = 2 - \frac{(8 - 4 - 5)}{(12 - 2)} = 2 - \frac{-1}{10} = 2.1.$$

Iteration 2:

$$x_2 = 2.1 - \frac{(2.1^3 - 2(2.1) - 5)}{3(2.1)^2 - 2} \approx 2.094.$$

Thus, the root is approximately $x \approx 2.094$.

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19. Find the maximum area of a rectangle inscribed in the parabola $y = 4 - x^2$.

Solution:
Let rectangle have width $2x$ and height $y = 4 - x^2$.
Area:

$$A = 2x(4 - x^2) = 8x - 2x^3.$$

First derivative:

$$A'(x) = 8 - 6x^2.$$

Setting $A'(x) = 0$,

$$6x^2 = 8 \Rightarrow x = \sqrt{\frac{4}{3}}.$$

Second derivative:

$$A''(x) = -12x.$$

At $x = \sqrt{4/3}$, $A''(x) < 0$, so it’s a maximum.

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20. Evaluate $\lim\limits_{x \to \infty} \frac{x}{\ln x}$ using L'Hôpital’s Rule.

Solution:
Since $\frac{x}{\ln x}$ is an $\frac{\infty}{\infty}$ form, apply L'Hôpital's Rule:

$$\lim\limits_{x \to \infty} \frac{x}{\ln x} = \lim\limits_{x \to \infty} \frac{1}{(1/x)} = \lim\limits_{x \to \infty} x = \infty.$$

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21. Find the radius of curvature of $y = x^3$ at $x = 1$.

Solution:
Radius of curvature formula:

$$R = \frac{(1 + (y')^2)^{3/2}}{|y''|}.$$

First derivative:

$$y' = 3x^2.$$

Second derivative:

$$y'' = 6x.$$

At $x = 1$,

$$R = \frac{(1 + 9)^{3/2}}{|6|} = \frac{10^{3/2}}{6}.$$

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22. Find the length of the curve $y = \frac{1}{3}x^3$ from $x = 0$ to $x = 1$.

Solution:
Arc length formula:

$$L = \int_{a}^{b} \sqrt{1 + (y')^2} \, dx.$$ $$y' = x^2, \quad (y')^2 = x^4.$$ $$L = \int_{0}^{1} \sqrt{1 + x^4} \, dx.$$

Approximating using binomial expansion:

$$L \approx \int_0^1 (1 + \frac{1}{2}x^4)dx = x + \frac{1}{10}x^5 \Big|_0^1 = 1 + \frac{1}{10} = 1.1.$$

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23. Find the minimum value of $f(x) = x + \frac{1}{x}$ for $x > 0$.

Solution:

$$f'(x) = 1 - \frac{1}{x^2}.$$

Setting $f'(x) = 0$,

$$1 = \frac{1}{x^2} \Rightarrow x = 1.$$

Second derivative:

$$f''(x) = \frac{2}{x^3}.$$

Since $f''(1) > 0$, minimum value is $f(1) = 2$.

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24. Prove that $f(x) = x^3 + x + 1$ has exactly one real root.

Solution:
Since $f(x)$ is a cubic, it has at least one real root.
First derivative:

$$f'(x) = 3x^2 + 1 > 0 \quad \forall x.$$

Since $f'(x) > 0$, $f(x)$ is strictly increasing, implying a unique real root.

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25. Prove that the function $f(x) = x^5 - 5x + 1$ has at least three real roots.

Solution:
By Rolle’s Theorem, if $f(a) = f(b)$, then $f'(c) = 0$ in $(a, b)$.
First derivative:

$$f'(x) = 5x^4 - 5.$$

Setting $f'(x) = 0$,

$$5(x^4 - 1) = 0 \Rightarrow x = \pm1.$$

Since $f'(x)$ has two real roots, $f(x)$ has at least three real roots by the Intermediate Value Theorem.

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This completes 25 solved problems on Differentiation and Application of Derivatives with increasing difficulty.