12th Grade Mathematics: Differentiation Formulae and Practice Problems
Key Differentiation Formulae
- Power Rule: $\frac{d}{dx}(x^n) = nx^{n-1}$
- Constant Rule: $\frac{d}{dx}(c) = 0$
- Product Rule: $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$
- Quotient Rule: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$
- Chain Rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$
- Trigonometric Functions:
- $\frac{d}{dx}(\sin x) = \cos x$
- $\frac{d}{dx}(\cos x) = -\sin x$
- $\frac{d}{dx}(\tan x) = \sec^2 x$
- $\frac{d}{dx}(\cot x) = -\csc^2 x$
- $\frac{d}{dx}(\sec x) = \sec x \tan x$
- $\frac{d}{dx}(\csc x) = -\csc x \cot x$
- Exponential Functions: $\frac{d}{dx}(e^x) = e^x$, $\frac{d}{dx}(a^x) = a^x \ln a$
- Logarithmic Functions: $\frac{d}{dx}(\ln x) = \frac{1}{x}$, $\frac{d}{dx}(\log_a x) = \frac{1}{x \ln a}$
- Inverse Trigonometric Functions:
- $\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}}$
- $\frac{d}{dx}(\cos^{-1} x) = -\frac{1}{\sqrt{1-x^2}}$
- $\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}$
Problem 1: Basic Polynomial Differentiation
Differentiate $y = 4x^5 - 3x^3 + 7x - 2$ with respect to $x$.
Solution:
Using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$ and the constant rule:
$$\frac{dy}{dx} = \frac{d}{dx}(4x^5) - \frac{d}{dx}(3x^3) + \frac{d}{dx}(7x) - \frac{d}{dx}(2)$$
$$\frac{dy}{dx} = 4(5x^4) - 3(3x^2) + 7(1) - 0$$
$$\frac{dy}{dx} = 20x^4 - 9x^2 + 7$$
Problem 2: Product Rule
Find the derivative of $y = x^2 \sin x$.
Solution:
Let $u = x^2$ and $v = \sin x$. Apply the product rule: $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$
Here, $\frac{du}{dx} = 2x$ and $\frac{dv}{dx} = \cos x$.
$$\frac{dy}{dx} = x^2 \cdot \frac{d}{dx}(\sin x) + \sin x \cdot \frac{d}{dx}(x^2)$$
$$\frac{dy}{dx} = x^2 \cos x + (\sin x)(2x)$$
$$\frac{dy}{dx} = x^2 \cos x + 2x \sin x$$
$$\frac{dy}{dx} = x(x \cos x + 2 \sin x)$$
Problem 3: Quotient Rule
Differentiate $y = \frac{e^x}{x^2}$ with respect to $x$.
Solution:
Let $u = e^x$ and $v = x^2$. Apply the quotient rule: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$
Here, $\frac{du}{dx} = e^x$ and $\frac{dv}{dx} = 2x$.
$$\frac{dy}{dx} = \frac{x^2 \cdot \frac{d}{dx}(e^x) - e^x \cdot \frac{d}{dx}(x^2)}{(x^2)^2}$$
$$\frac{dy}{dx} = \frac{x^2 e^x - e^x(2x)}{x^4}$$
Factor out $x e^x$ from the numerator:
$$\frac{dy}{dx} = \frac{x e^x (x - 2)}{x^4} = \frac{e^x (x - 2)}{x^3}$$
Problem 4: Chain Rule
Find $\frac{dy}{dx}$ if $y = \cos(x^3 + 5x)$.
Solution:
Let $u = x^3 + 5x$, then $y = \cos(u)$. Apply the chain rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$
$$\frac{dy}{du} = -\sin(u)$$
$$\frac{du}{dx} = 3x^2 + 5$$
$$\frac{dy}{dx} = -\sin(x^3 + 5x) \cdot (3x^2 + 5)$$
$$\frac{dy}{dx} = -(3x^2 + 5)\sin(x^3 + 5x)$$
Problem 5: Implicit Differentiation
Find $\frac{dy}{dx}$ for the equation $x^2 + y^2 = 25$.
Solution:
Differentiate both sides with respect to $x$:
$$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25)$$
$$2x + 2y \frac{dy}{dx} = 0$$
Solve for $\frac{dy}{dx}$:
$$2y \frac{dy}{dx} = -2x$$
$$\frac{dy}{dx} = -\frac{2x}{2y} = -\frac{x}{y}$$
Problem 6: Parametric Differentiation
If $x = a \cos t$ and $y = b \sin t$, find $\frac{dy}{dx}$.
Solution:
First, find the derivatives of $x$ and $y$ with respect to the parameter $t$:
$$\frac{dx}{dt} = -a \sin t$$
$$\frac{dy}{dt} = b \cos t$$
Now, apply the parametric formula $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$:
$$\frac{dy}{dx} = \frac{b \cos t}{-a \sin t} = -\frac{b}{a} \cot t$$
Problem 7: Logarithmic Differentiation
Differentiate $y = x^x$ with respect to $x$.
Solution:
Take the natural logarithm ($\ln$) of both sides:
$$\ln y = \ln(x^x)$$
$$\ln y = x \ln x$$
Differentiate both sides implicitly with respect to $x$ (use product rule on the right):
$$\frac{1}{y} \frac{dy}{dx} = x \cdot \frac{d}{dx}(\ln x) + \ln x \cdot \frac{d}{dx}(x)$$
$$\frac{1}{y} \frac{dy}{dx} = x \left(\frac{1}{x}\right) + \ln x (1)$$
$$\frac{1}{y} \frac{dy}{dx} = 1 + \ln x$$
$$\frac{dy}{dx} = y(1 + \ln x)$$
Substitute back $y = x^x$:
$$\frac{dy}{dx} = x^x(1 + \ln x)$$
Problem 8: Derivative of an Inverse Trigonometric Function
Find the derivative of $y = \sin^{-1}(2x)$.
Solution:
Use the chain rule along with the formula $\frac{d}{dx}(\sin^{-1} u) = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}$.
Here, $u = 2x$, so $\frac{du}{dx} = 2$.
$$\frac{dy}{dx} = \frac{1}{\sqrt{1 - (2x)^2}} \cdot \frac{d}{dx}(2x)$$
$$\frac{dy}{dx} = \frac{1}{\sqrt{1 - 4x^2}} \cdot 2 = \frac{2}{\sqrt{1 - 4x^2}}$$
Problem 9: Derivative Involving Exponential and Chain Rules
Differentiate $y = e^{\tan x}$.
Solution:
Use the chain rule. The outer function is $e^u$ and the inner function is $u = \tan x$.
$$\frac{dy}{dx} = e^{\tan x} \cdot \frac{d}{dx}(\tan x)$$
$$\frac{dy}{dx} = e^{\tan x} \cdot \sec^2 x$$
Problem 10: Higher Order Derivative
Find the second derivative, $\frac{d^2y}{dx^2}$, of the function $y = x^3 \ln x$.
Solution:
First, find the first derivative $\frac{dy}{dx}$ using the product rule:
$$\frac{dy}{dx} = x^3 \cdot \frac{d}{dx}(\ln x) + \ln x \cdot \frac{d}{dx}(x^3)$$
$$\frac{dy}{dx} = x^3 \left(\frac{1}{x}\right) + (\ln x)(3x^2) = x^2 + 3x^2 \ln x$$
Now, differentiate again to find the second derivative:
$$\frac{d^2y}{dx^2} = \frac{d}{dx}(x^2) + \frac{d}{dx}(3x^2 \ln x)$$
Apply the product rule again to the second term:
$$\frac{d^2y}{dx^2} = 2x + \left[3x^2 \cdot \frac{1}{x} + (\ln x)(6x)\right]$$
$$\frac{d^2y}{dx^2} = 2x + 3x + 6x \ln x$$
$$\frac{d^2y}{dx^2} = 5x + 6x \ln x = x(5 + 6 \ln x)$$
