Differentiation Techniques in Calculus: Study Notes and Applications

Differentiation Techniques in Calculus

1. Basic Differentiation of Functions

Differentiation is a fundamental concept in calculus, used to determine the rate at which a function changes. The derivative of a function, denoted as $y'(x)$ or $\frac{dy}{dx}$, provides the slope of the tangent line at any point on the curve.

• Key Point: To differentiate a function, apply the rules for each term separately, including the power rule, chain rule, and derivative of logarithmic and exponential functions.

• Example: For $y = \ln(x + e^x) + 2x^7 + 5x^{12} + 6^x$:

Derivative:

• $\frac{d}{dx}[\ln(x + e^x)] = \frac{1}{x + e^x} \cdot (1 + e^x)$ (by chain rule)

• $\frac{d}{dx}[2x^7] = 14x^6$

• $\frac{d}{dx}[5x^{12}] = 60x^{11}$

• $\frac{d}{dx}[6^x] = 6^x \ln 6$

Combined:

• $y'(x) = \frac{1 + e^x}{x + e^x} + 14x^6 + 60x^{11} + 6^x \ln 6$

2. Implicit Differentiation

Implicit differentiation is used when a function is defined implicitly rather than explicitly. This technique is essential when $y$ cannot be easily isolated.

• Key Point: Differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$ and applying the chain rule where necessary.

• Example: Given $\sin y = x \cos y - 2$, find $\frac{dy}{dx}$.

Steps:

• Differentiate both sides: $\cos y \frac{dy}{dx} = \cos y - x \sin y \frac{dy}{dx}$

• Collect $\frac{dy}{dx}$ terms and solve:

$\cos y \frac{dy}{dx} + x \sin y \frac{dy}{dx} = \cos y$

$\frac{dy}{dx}(\cos y + x \sin y) = \cos y$

$\frac{dy}{dx} = \frac{\cos y}{\cos y + x \sin y}$

3. Logarithmic Differentiation

Logarithmic differentiation is useful for differentiating functions involving products, quotients, or powers where direct differentiation is complex. It involves taking the natural logarithm of both sides before differentiating.

• Key Point: Use properties of logarithms to simplify the function before differentiating.

• Example: For $y = \ln\left(\sqrt{\frac{x^2 \cos x}{1 + e^{2x}}}\right)$:

Steps:

• Rewrite using logarithm properties:

$y = \frac{1}{2} \ln\left(\frac{x^2 \cos x}{1 + e^{2x}}\right)$

$= \frac{1}{2} [\ln(x^2) + \ln(\cos x) - \ln(1 + e^{2x})]$

• Differentiating term by term:

$\frac{dy}{dx} = \frac{1}{2} \left[ \frac{2}{x} - \tan x - \frac{2e^{2x}}{1 + e^{2x}} \right]$

4. Logarithmic Differentiation for Exponential Equations

When both the base and exponent are variables, logarithmic differentiation simplifies the process.

• Key Point: Take the natural logarithm of both sides, differentiate, and solve for $\frac{dy}{dx}$.

• Example: Given $x^{\sin y} = \ln y$:

Steps:

• Take $\ln$ of both sides: $\sin y \ln x = \ln(\ln y)$

• Differentiating both sides with respect to $x$:

$\cos y \frac{dy}{dx} \ln x + \sin y \frac{1}{x} = \frac{1}{\ln y} \frac{1}{y} \frac{dy}{dx}$

• Collect $\frac{dy}{dx}$ terms and solve as needed.

Additional info: The full solution would require isolating $\frac{dy}{dx}$, but the process above demonstrates the method.

5. Differentiation of Inverse Trigonometric Functions

Inverse trigonometric functions require special differentiation rules. The derivatives often involve algebraic expressions and chain rule applications.

• Key Point: Use the standard derivatives for $\sin^{-1}x$, $\cos^{-1}x$, and $\tan^{-1}x$:

Function	Derivative
$\sin^{-1}x$	$\frac{1}{\sqrt{1 - x^2}}$
$\cos^{-1}x$	$-\frac{1}{\sqrt{1 - x^2}}$
$\tan^{-1}x$	$\frac{1}{1 + x^2}$

• Example: For $y = 2\sin^{-1}(e^x + x) - 3\cos^{-1}(9) + 4\tan^{-1}(2x - 5)$:

Derivative:

• $\frac{d}{dx}[2\sin^{-1}(e^x + x)] = 2 \cdot \frac{1}{\sqrt{1 - (e^x + x)^2}} \cdot (e^x + 1)$

• $\frac{d}{dx}[-3\cos^{-1}(9)] = 0$ (since $\cos^{-1}(9)$ is constant)

• $\frac{d}{dx}[4\tan^{-1}(2x - 5)] = 4 \cdot \frac{1}{1 + (2x - 5)^2} \cdot 2$

Combined:

• $y'(x) = \frac{2(e^x + 1)}{\sqrt{1 - (e^x + x)^2}} + \frac{8}{1 + (2x - 5)^2}$

6. Related Rates

Related rates problems involve finding the rate of change of one quantity in terms of the rate of change of another, often using implicit differentiation.

• Key Point: Differentiate the given relationship with respect to time $t$, then substitute known values to solve for the desired rate.

• Example: Given $x^2 y^2 = 4/27$, $\frac{dy}{dt} = 1/2$, and $x = 2$, find $\frac{dx}{dt}$ when $x = 2$.

Steps:

• Differentiate both sides with respect to $t$:

$2x y^2 \frac{dx}{dt} + 2x^2 y \frac{dy}{dt} = 0$

• Solve for $\frac{dx}{dt}$:

$2x y^2 \frac{dx}{dt} = -2x^2 y \frac{dy}{dt}$

$\frac{dx}{dt} = -\frac{x^2 y}{x y^2} \frac{dy}{dt} = -\frac{x y}{y^2} \frac{dy}{dt}$

• Substitute $x = 2$, $\frac{dy}{dt} = 1/2$, and solve for $y$ using $x^2 y^2 = 4/27$:

$4 y^2 = 4/27 \implies y^2 = 1/27 \implies y = 1/\sqrt{27}$

$\frac{dx}{dt} = -\frac{2 \cdot 1/\sqrt{27}}{1/27} \cdot \frac{1}{2} = -27 \cdot \frac{1}{\sqrt{27}} \cdot \frac{1}{2}$

Additional info: The final numerical value can be computed as needed.