Understanding Derivative Examples in Calculus

The derivative of a function with respect to its variable measures the instantaneous rate of change at each point in the domain. Derivative calculations provide the foundational tools in calculus for studying local behavior, tangency, monotonicity, extrema, and concavity. The following exposition develops and illustrates examples of derivatives, employing explicit step-by-step computations for algebraic, trigonometric, and implicit forms as encountered in calculus.

Algebraic Computation of Derivatives: Stepwise Examples

Given: $f(x) = 4x^5 - 3x^3 + 7x^2 - 6x + 9$.

Rewrite as a sum of terms for component-wise differentiation: $$ f(x) = 4x^5 - 3x^3 + 7x^2 - 6x + 9 $$ Apply the power rule for each term separately.

The derivative of $4x^5$ with respect to $x$ is: $$ \dfrac{d}{dx}(4x^5) = 4 \cdot 5 x^{5-1} = 20x^4 $$ The derivative of $-3x^3$ is: $$ \dfrac{d}{dx}(-3x^3) = -3 \cdot 3 x^{3-1} = -9x^2 $$ The derivative of $7x^2$ is: $$ \dfrac{d}{dx}(7x^2) = 7 \cdot 2 x^{2-1} = 14x $$ The derivative of $-6x$ is: $$ \dfrac{d}{dx}(-6x) = -6 $$ The derivative of $9$ is: $$ \dfrac{d}{dx}(9) = 0 $$ Combine all results: $$ f'(x) = 20x^4 - 9x^2 + 14x - 6 $$

Result: $f'(x) = 20x^4 - 9x^2 + 14x - 6$.

Explicit Derivative of Trigonometric Functions: Illustrative Calculation

Given: $g(x) = \sin(2x) + \cos x$.

Differentiate each term separately. Begin with $\sin(2x)$ by applying the chain rule. $$ \dfrac{d}{dx}(\sin(2x)) = \cos(2x)\cdot \dfrac{d}{dx}(2x) = \cos(2x)\cdot 2 = 2\cos(2x) $$ For $\cos x$, use the standard derivative: $$ \dfrac{d}{dx}(\cos x) = -\sin x $$ Thus, $$ g'(x) = 2\cos(2x) - \sin x $$

Result: $g'(x) = 2\cos(2x) - \sin x$.

Derivative of Inverse Trigonometric Function: Methodical Evaluation

Given: $h(x) = \arccos x$.

The formula for the derivative of $\arccos x$ is: $$ \dfrac{d}{dx}(\arccos x) = -\dfrac{1}{\sqrt{1 - x^2}} $$ Therefore, $$ h'(x) = -\dfrac{1}{\sqrt{1 - x^2}} $$

Result: $h'(x) = -\dfrac{1}{\sqrt{1 - x^2}}$.

Second Derivative Calculation for Polynomial Functions

Given: $y = x^4 - 2x^3 + 3x^2$.

First, compute the first derivative: $$ \dfrac{dy}{dx} = \dfrac{d}{dx}(x^4) - \dfrac{d}{dx}(2x^3) + \dfrac{d}{dx}(3x^2) $$ Use the power rule for each term: $$ \dfrac{d}{dx}(x^4) = 4x^{3} $$ $$ \dfrac{d}{dx}(2x^3) = 2 \cdot 3 x^{2} = 6x^2 $$ $$ \dfrac{d}{dx}(3x^2) = 3 \cdot 2 x^{1} = 6x $$ Combine: $$ \dfrac{dy}{dx} = 4x^3 - 6x^2 + 6x $$ Now, differentiate again to obtain the second derivative: $$ \dfrac{d^2 y}{dx^2} = \dfrac{d}{dx}(4x^3) - \dfrac{d}{dx}(6x^2) + \dfrac{d}{dx}(6x) $$ $$ \dfrac{d}{dx}(4x^3) = 12x^2 $$ $$ \dfrac{d}{dx}(6x^2) = 12x $$ $$ \dfrac{d}{dx}(6x) = 6 $$ Therefore: $$ \dfrac{d^2 y}{dx^2} = 12x^2 - 12x + 6 $$

Result: $\dfrac{d^2 y}{dx^2} = 12x^2 - 12x + 6$.

Use of Product and Quotient Rule in Differentiation

Given: $f(x) = (2x + 5)(x^3 - 4)$.

Let $u(x) = 2x + 5$ and $v(x) = x^3 - 4$. The product rule states: $$ \dfrac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) $$ Compute $u'(x)$ and $v'(x)$: $$ u'(x) = \dfrac{d}{dx}(2x + 5) = 2 $$ $$ v'(x) = \dfrac{d}{dx}(x^3 - 4) = 3x^2 $$ Substitute: $$ f'(x) = 2 \cdot (x^3 - 4) + (2x + 5) \cdot 3x^2 $$ Expand: $$ f'(x) = 2x^3 - 8 + 6x^3 + 15x^2 $$ Combine like terms: $$ f'(x) = (2x^3 + 6x^3) + 15x^2 - 8 = 8x^3 + 15x^2 - 8 $$

Result: $f'(x) = 8x^3 + 15x^2 - 8$.

For comprehensive rules such as the product, chain, and quotient rule, refer also to Differentiability of Composite Functions.

Implicit Differentiation: Explicit Solution

Given: $x^2 + y^2 = 49$.

Differentiate both sides with respect to $x$. Treat $y$ as a function of $x$. $$ \dfrac{d}{dx}(x^2 + y^2) = \dfrac{d}{dx}(49) $$ By linearity: $$ \dfrac{d}{dx}(x^2) + \dfrac{d}{dx}(y^2) = 0 $$ Differentiate $x^2$: $$ \dfrac{d}{dx}(x^2) = 2x $$ Differentiate $y^2$ using the chain rule: $$ \dfrac{d}{dx}(y^2) = 2y \dfrac{dy}{dx} $$ Thus: $$ 2x + 2y \dfrac{dy}{dx} = 0 $$ Subtract $2x$: $$ 2y \dfrac{dy}{dx} = -2x $$ Divide both sides by $2y$: $$ \dfrac{dy}{dx} = -\dfrac{x}{y} $$

Result: $\dfrac{dy}{dx} = -\dfrac{x}{y}$.

Differentiation of Composite Trigonometric Functions: Chain Rule Application

Given: $y = \sin(3x)$.

Express as $y = \sin(u)$ where $u = 3x$. Then, $\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$. $$ \dfrac{dy}{du} = \cos(u) $$ $$ \dfrac{du}{dx} = 3 $$ Thus: $$ \dfrac{dy}{dx} = \cos(3x) \cdot 3 = 3\cos(3x) $$

Result: $\dfrac{dy}{dx} = 3\cos(3x)$.

For further worked problems in calculus, see Differential Calculus.

Sample Problems on Derivatives: Practice Set

Differentiate the following with respect to $x$:

(i) $f(x) = e^{2x} + \ln x$

(ii) $y = \cos^2 x + \sin^2 x$

(iii) $f(x) = \dfrac{x^2 + 1}{x^3 - x}$

(iv) $y = \arctan(2x)$

(v) $x^2 + xy + y^2 = 10$ (Use implicit differentiation)

Detailed solutions to these examples, incorporating all necessary derivative rules, are recommended for exam preparation and can be found in advanced practice sections of Derivative Examples.