Class 12 Maths Chapter 6 Application of Derivatives – Exercise 6.1 NCERT Solutions

Introduction

Exercise 6.1 focuses on rate of change of quantities using derivatives. Students learn how to apply differentiation to real‑life problems such as velocity, growth, decay, and geometric applications. This exercise builds the foundation for practical applications of calculus.

Formulas Used

1. Rate of Change: If $y=f(x)$, then rate of change of $y$ w.r.t. $x$ is:

$$\frac{dy}{dx}$$

2. Velocity and Acceleration:

$$v=\frac{ds}{dt},a=\frac{dv}{dt}=\frac{d^{2}s}{d{t}^2}$$

3. Derivative in Geometry: For radius $r$,

$$\frac{dA}{dr}\text{(rate of change of area)},\frac{dV}{dr}\text{(rate of change of volume)}$$

Students Frequently Make Mistakes

• Forgetting to apply chain rule in composite functions.

• Misinterpreting physical meaning of derivative.

• Errors in unit conversion while applying rate of change.

• Skipping second derivative for acceleration problems.

• Confusing instantaneous rate with average rate.

NCERT Questions with Step‑by‑Step Solutions (10)

Q1. If $y=x^3$, find rate of change of $y$ w.r.t. $x$ at $x=2$.

$$\frac{dy}{dx}=3x^2,\frac{dy}{dx}{\mid }_{x=2}=12$$

Q2. If $y=\sqrt{x}$, find rate of change at $x=4$.

$$\frac{dy}{dx}=\frac{1}{2\sqrt{x}},\frac{dy}{dx}{\mid }_{x=4}=\frac{1}{4}$$

Q3. If $y=\ln x$, find rate of change at $x=e$.

$$\frac{dy}{dx}=\frac{1}{x},\frac{dy}{dx}{\mid }_{x=e}=\frac{1}{e}$$

Q4. If $y=\sin x$, find rate of change at $x=\pi \mathrm{/}4$.

$$\frac{dy}{dx}=\cos x,\frac{dy}{dx}{\mid }_{x=\pi \mathrm{/}4}=\frac{\sqrt{2}}{2}$$

Q5. If $y=\cos x$, find rate of change at $x=\pi \mathrm{/}3$.

$$\frac{dy}{dx}=-\sin x,\frac{dy}{dx}{\mid }_{x=\pi \mathrm{/}3}=-\frac{\sqrt{3}}{2}$$

Q6. If $s=t^2+3t$, find velocity at $t=2$.

$$v=\frac{ds}{dt}=2t+3,v(2)=7$$

Q7. If $s=t^3$, find acceleration at $t=2$.

$$v=\frac{ds}{dt}=3t^2,a=\frac{dv}{dt}=6t,a(2)=12$$

Q8. If area of circle $A=\pi r^2$, find rate of change of area w.r.t. radius at $r=7$.

$$\frac{dA}{dr}=2\pi r,\frac{dA}{dr}{\mid }_{r=7}=14\pi$$

Q9. If volume of sphere $V=\frac{4}{3}\pi r^3$, find rate of change of volume w.r.t. radius at $r=3$.

$$\frac{dV}{dr}=4\pi r^2,\frac{dV}{dr}{\mid }_{r=3}=36\pi$$

Q10. If $y=e^x$, find rate of change at $x=0$.

$$\frac{dy}{dx}=e^x,\frac{dy}{dx}{\mid }_{x=0}=1$$

FAQs (10)

FAQ1. What is rate of change? Derivative of one variable w.r.t. another.

FAQ2. What is velocity in calculus? First derivative of displacement w.r.t. time.

FAQ3. What is acceleration in calculus? Second derivative of displacement w.r.t. time.

FAQ4. What is derivative of area of circle? $\frac{dA}{dr}=2\pi r$.

FAQ5. What is derivative of volume of sphere? $\frac{dV}{dr}=4\pi r^2$.

FAQ6. What is instantaneous rate of change? Derivative at a specific point.

FAQ7. What is average rate of change? $\frac{f(b)-f(a)}{b-a}$.

FAQ8. Why use derivatives in physics? To calculate velocity, acceleration, growth, decay.

FAQ9. Why check domain restrictions? Functions like $\ln x$ defined only for $x>0$.

FAQ10. Why is Exercise 6.1 important? It builds foundation for real‑life applications of derivatives.

Conclusion

Exercise 6.1 has 10 solved questions and 10 FAQs that strengthen your understanding of rate of change using derivatives. This begins the Applications of Derivatives chapter in Class 12 Maths.

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