6.2 Derivatives of Trigonometric, Exponential & Logarithmic Functions

Derivatives of Trigonometric, Exponential & Logarithmic Functions

Transcendental functions—encompassing trigonometric, inverse trigonometric, exponential, and logarithmic families—constitute the core of advanced calculus questions on the UET ECAT. These functions frequently appear in composite arrangements requiring simultaneous deployment of the chain rule, product rule, and logarithmic differentiation. A precise command of standard derivative formulas, their domain restrictions, and high-speed algebraic identities is essential to achieving maximum speed and accuracy on test day.

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1. Derivatives of the Six Circular Trigonometric Functions

Analytical Derivations via First Principles

All trigonometric derivatives are derived from first principles using the fundamental trigonometric limit:

$\lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1 \quad (\theta \text{ in radians})$

Critical Requirement: In all calculus formulas involving trigonometric functions, the angle $x$ must be measured in radians. If an ECAT question gives the angle in degrees ($x^\circ$), convert it first: $x^\circ = \frac{\pi x}{180}$ radians. For example: $\frac{d}{dx}(\sin x^\circ) = \frac{d}{dx}\left(\sin\left(\frac{\pi x}{180}\right)\right) = \frac{\pi}{180}\cos\left(\frac{\pi x}{180}\right) = \frac{\pi}{180}\cos x^\circ$ Overlooking the factor $\frac{\pi}{180}$ is one of the most common traps set by examiners!

The Six Standard Formulas

1. Sine: $\frac{d}{dx}(\sin x) = \cos x$

2. Cosine: $\frac{d}{dx}(\cos x) = -\sin x$

3. Tangent: (via quotient rule on $\frac{\sin x}{\cos x}$): $\frac{d}{dx}(\tan x) = \frac{(\cos x)(\cos x) - (\sin x)(-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x$

4. Cotangent: $\frac{d}{dx}(\cot x) = -\csc^2 x$

5. Secant: $\frac{d}{dx}(\sec x) = \sec x \tan x$

6. Cosecant: $\frac{d}{dx}(\csc x) = -\csc x \cot x$

The "Co-Function" Negative Sign Mnemonic

Notice the symmetry: all three co-functions ($\cos x$, $\cot x$, and $\csc x$) have derivatives prefixed with an explicit negative sign. Non-co functions ($\sin x$, $\tan x$, $\sec x$) produce strictly positive-signed formulas.

Chain Rule Generalization for Trigonometric Functions

If $u = u(x)$ is a differentiable function of $x$:

• $\frac{d}{dx}[\sin(u)] = \cos(u) \cdot u'$
• $\frac{d}{dx}[\cos(u)] = -\sin(u) \cdot u'$
• $\frac{d}{dx}[\tan(u)] = \sec^2(u) \cdot u'$
• $\frac{d}{dx}[\cot(u)] = -\csc^2(u) \cdot u'$
• $\frac{d}{dx}[\sec(u)] = \sec(u)\tan(u) \cdot u'$
• $\frac{d}{dx}[\csc(u)] = -\csc(u)\cot(u) \cdot u'$

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2. Derivatives of Inverse Trigonometric Functions

Inverse trigonometric functions are defined on restricted domains to ensure injectivity. Their derivatives map inverse transcendental functions into purely algebraic rational and radical expressions.

Derivation of $\frac{d}{dx}(\arcsin x)$

Let $y = \arcsin x = \sin^{-1} x$, where $x \in (-1, 1)$ and $y \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

1. Invert to explicit trigonometric form: $x = \sin y$.
2. Differentiate implicitly with respect to $x$: $1 = \cos y \frac{dy}{dx}$.
3. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{1}{\cos y}$.
4. Express in terms of $x$: Since $y \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, $\cos y > 0$, so $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$.

$\frac{d}{dx}(\arcsin x) = \frac{1}{\sqrt{1 - x^2}}, \quad |x| < 1$

Complete Set of Inverse Trigonometric Derivatives

Complementary Symmetry

Observe that complementary pairs add up to a constant: $\arcsin x + \arccos x = \frac{\pi}{2}$ and $\arctan x + \text{arccot } x = \frac{\pi}{2}$. Differentiating both sides of $\arcsin x + \arccos x = \frac{\pi}{2}$ gives $\frac{d}{dx}(\arcsin x) + \frac{d}{dx}(\arccos x) = 0 \implies \frac{d}{dx}(\arccos x) = -\frac{d}{dx}(\arcsin x)$.

High-Speed Trigonometric Substitution Shortcuts

ECAT questions often ask for derivatives of expressions that appear formidable at first glance, but collapse under standard half-angle or multiple-angle identities:

• $\frac{d}{dx}\left[\arcsin(2x\sqrt{1-x^2})\right]$: Substitute $x = \sin\theta \implies 2x\sqrt{1-x^2} = \sin(2\theta)$. For $|x| \le \frac{1}{\sqrt{2}}$ (so that $2\theta$ stays inside $[-\frac{\pi}{2}, \frac{\pi}{2}]$), the function reduces to $2\theta = 2\arcsin x$, whose derivative is $\frac{2}{\sqrt{1-x^2}}$ on $|x| < \frac{1}{\sqrt{2}}$. For $\frac{1}{\sqrt{2}} < |x| < 1$ the function equals $\pm\pi - 2\arcsin x$ and the derivative becomes $-\frac{2}{\sqrt{1-x^2}}$.
• $\frac{d}{dx}\left[\arctan\left(\frac{2x}{1-x^2}\right)\right]$: Substitute $x = \tan\theta \implies \frac{2\tan\theta}{1-\tan^2\theta} = \tan(2\theta)$. The expression equals $2\arctan x$ for $|x| < 1$ (and differs from it by the constant $\pm\pi$ when $|x| > 1$), so the derivative is $\frac{2}{1+x^2}$ for every $x \neq \pm 1$.
• $\frac{d}{dx}\left[\arcsin(3x - 4x^3)\right]$: Substitute $x = \sin\theta \implies 3\sin\theta - 4\sin^3\theta = \sin(3\theta)$. For $|x| \le \frac{1}{2}$ the function simplifies to $3\arcsin x$, giving derivative $\frac{3}{\sqrt{1-x^2}}$ on $|x| < \frac{1}{2}$ (outside that interval $3\theta$ leaves the principal range and the simplification fails).

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3. Derivatives of Exponential Functions

The Natural Exponential Function

The number $e \approx 2.71828$ is defined analytically such that $\lim_{h \to 0} \frac{e^h - 1}{h} = 1$. Consequently, the natural exponential function $f(x) = e^x$ is unique because its instantaneous rate of change equals its instantaneous value:

$\frac{d}{dx}(e^x) = e^x$

Applying the chain rule for $y = e^{u(x)}$: $\frac{d}{dx}(e^{u(x)}) = e^{u(x)} \cdot u'(x)$

General Exponential Functions ($a^x$)

For an arbitrary positive base $a > 0$ with $a \neq 1$, rewrite $a^x$ using the natural base identity $a^x = e^{\ln(a^x)} = e^{x \ln a}$:

$\frac{d}{dx}(a^x) = \frac{d}{dx}(e^{x \ln a}) = e^{x \ln a} \cdot \ln a = a^x \ln a$

Applying the chain rule for $y = a^{u(x)}$: $\frac{d}{dx}(a^{u(x)}) = a^{u(x)} \ln a \cdot u'(x)$

ECAT Key Fact: Notice that $\frac{d}{dx}(e^x) = e^x \ln e = e^x(1) = e^x$. The natural base $e$ is simply the special case where the scaling factor $\ln a$ equals unity.

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4. Derivatives of Logarithmic Functions

The Natural Logarithm ($\ln x$)

Let $y = \ln x$ ($x > 0$). Inverting gives $x = e^y$. Differentiating implicitly w.r.t $x$ yields $1 = e^y \frac{dy}{dx} = x \frac{dy}{dx}$, which provides:

$\frac{d}{dx}(\ln x) = \frac{1}{x}, \quad x > 0$

For real non-zero values, $y = \ln |x|$ ($x \neq 0$): $\frac{d}{dx}(\ln |x|) = \frac{1}{x}$

With the chain rule for $y = \ln(u(x))$: $\frac{d}{dx}[\ln(u(x))] = \frac{u'(x)}{u(x)}$ This expression $\frac{u'(x)}{u(x)}$ is termed the logarithmic derivative of $u(x)$.

General Logarithmic Functions ($\log_a x$)

Using the change of base formula $\log_a x = \frac{\ln x}{\ln a}$:

$\frac{d}{dx}(\log_a x) = \frac{d}{dx}\left(\frac{\ln x}{\ln a}\right) = \frac{1}{x \ln a}$

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5. Logarithmic Differentiation Technique

Logarithmic differentiation is an indispensable problem-solving tool on the ECAT when facing:

1. Expressions with multiple factors multiplied or divided together: $y = \frac{f_1(x) f_2(x) \dots}{g_1(x) g_2(x) \dots}$.
2. Power-tower functions where both the base and exponent are functions of $x$: $y = [f(x)]^{g(x)}$.

Derivation for Power-Tower Form $y = [f(x)]^{g(x)}$

Take the natural logarithm of both sides: $\ln y = \ln\left([f(x)]^{g(x)}\right) = g(x) \ln(f(x))$

Differentiate implicitly with respect to $x$ using the product rule on the right-hand side: $\frac{1}{y} \frac{dy}{dx} = g'(x) \ln(f(x)) + g(x) \cdot \frac{f'(x)}{f(x)}$

Multiply through by $y = [f(x)]^{g(x)}$ to isolate $\frac{dy}{dx}$:

$\frac{dy}{dx} = [f(x)]^{g(x)} \left[ g'(x) \ln(f(x)) + \frac{g(x) f'(x)}{f(x)} \right]$

Canonical ECAT Results to Memorize

1. $y = x^x$: $\ln y = x \ln x \implies \frac{1}{y} y' = (1)\ln x + x\left(\frac{1}{x}\right) = 1 + \ln x$ $\frac{d}{dx}(x^x) = x^x(1 + \ln x)$

2. $y = x^{\sin x}$: $\frac{d}{dx}(x^{\sin x}) = x^{\sin x}\left[ \cos x \ln x + \frac{\sin x}{x} \right]$

3. $y = (\sin x)^x$: $\frac{d}{dx}\left((\sin x)^x\right) = (\sin x)^x [\ln(\sin x) + x \cot x]$

4. Rapid Product Simplification: If $y = \frac{(x+1)^3 (2x-5)^4}{\sqrt{3x^2+1}}$, take logarithms to break into additive components: $\ln y = 3\ln(x+1) + 4\ln(2x-5) - \frac{1}{2}\ln(3x^2+1)$ $\frac{y'}{y} = \frac{3}{x+1} + \frac{4(2)}{2x-5} - \frac{1}{2}\frac{6x}{3x^2+1} = \frac{3}{x+1} + \frac{8}{2x-5} - \frac{3x}{3x^2+1}$ This replaces pages of messy quotient and product expansions with a 2-line calculation.