derivatives lec1

Significance of Derivatives in Aptitude Tests

  • Exam Importance: The study of derivatives is a cornerstone for various aptitude examinations in Pakistan, including NED, NUST, FAST, DAWOOD, and the NTS. Mastery of this topic is essential for success in Class 11 and Class 12 (Intermediate) board exams.

  • Point Distribution (NED University): In the NED engineering entrance exam, the mathematics section consists of 25 marks. Historically, derivatives alone contribute between 2 to 3 multiple-choice questions (MCQs), making it a high-weightage chapter relative to its size.

  • Diagnostic of Student Failure: Approximately 80% of students in Karachi, Sindh, and across Pakistan struggle with derivatives primarily because they do not memorize or understand the fundamental formulas and properties. Without these, solving advanced calculus problems in competitive exams is impossible.

  • Pedagogical Approach: These notes are derived from a comprehensive analysis of past papers (NED, NUST, KU, and FAST). The teaching philosophy emphasizes understanding the logic behind formulas rather than rote memorization.

Basic Rules of Differentiation

  • The Power Formula: This is applied when a derivative is taken of a term raised to a power. The rule states: bring the power to the front (multiply), and subtract one from the original exponent. Finally, multiply by the derivative of the inner function.

    • Formula: ddx[xn]=nxn1ddx[x]\frac{d}{dx}[x^n] = n x^{n-1} \cdot \frac{d}{dx}[x]

    • Example: For 5x35x^3, the constant 55 remains. Bringing the power 33 down: 5×3x31=15x25 \times 3x^{3-1} = 15x^2.

    • Key Property: The derivative of variable xx with respect to xx is always 11 (dxdx=1\frac{dx}{dx} = 1), as is the derivative of yy with respect to yy (dydy=1\frac{dy}{dy} = 1).

Operation Formulas: Product and Quotient Rules

  • Multiplication Formula (Product Rule): Used when two distinct mathematical terms or functions are multiplying.

    • Procedure: Keep the first term as is and differentiate the second term. Add this to the product of the second term (kept as is) and the derivative of the first term.

    • Conceptual explanation: "Step aside and let the other work, then switch."

    • Example: ddx[xex]=xddx[ex]+exddx[x]\frac{d}{dx}[x e^x] = x \cdot \frac{d}{dx}[e^x] + e^x \cdot \frac{d}{dx}[x].

  • Division Formula (Quotient Rule): Used for terms in a fractional (numerator/denominator) format.

    • Procedure:

      1. Square the denominator term and place it at the bottom.

      2. In the numerator, write the original denominator term once.

      3. Multiply it by the derivative of the numerator function.

      4. Insert a MINUS sign (a common area for student error where they mistakenly use a plus).

      5. Multiply the original numerator term by the derivative of the denominator.

    • Formula: ddx[f(x)g(x)]=g(x)ddx[f(x)]f(x)ddx[g(x)][g(x)]2\frac{d}{dx}[\frac{f(x)}{g(x)}] = \frac{g(x) \cdot \frac{d}{dx}[f(x)] - f(x) \cdot \frac{d}{dx}[g(x)]}{[g(x)]^2}.

Logarithmic and Exponential Derivatives

  • Derivative of Natural Log (ln): The term ln\ln represents loge\log_e.

    • Formula: ddx[ln(f(x))]=1f(x)ddx[f(x)]\frac{d}{dx}[\ln(f(x))] = \frac{1}{f(x)} \cdot \frac{d}{dx}[f(x)]

    • Example: ddx[ln(x2)]=1x22x=2x\frac{d}{dx}[\ln(x^2)] = \frac{1}{x^2} \cdot 2x = \frac{2}{x}.

  • Common Exponential Derivative (af(x)a^{f(x)}): Applied when a numeric value (constant) is raised to a functional power.

    • Formula: ddx[af(x)]=af(x)ln(a)ddx[f(x)]\frac{d}{dx}[a^{f(x)}] = a^{f(x)} \cdot \ln(a) \cdot \frac{d}{dx}[f(x)]

    • Example: For 5sin(x)5^{\sin(x)}, the result is 5sin(x)ln(5)cos(x)5^{\sin(x)} \cdot \ln(5) \cdot \cos(x).

  • Natural Exponential Derivative (ef(x)e^{f(x)}):

    • Formula: ddx[ef(x)]=ef(x)ddx[f(x)]\frac{d}{dx}[e^{f(x)}] = e^{f(x)} \cdot \frac{d}{dx}[f(x)]

    • Example: For ex2e^{x^2}, the result is ex22xe^{x^2} \cdot 2x.

Mathematical Properties of Logarithms (ln)

  • Product to Sum Rule: If a log is applied to the product of two variables, it can be split into the sum of logs.

    • Rule: ln(a×b)=ln(a)+ln(b)\ln(a \times b) = \ln(a) + \ln(b)

  • Quotient to Difference Rule: If a log is applied to a division, it splits into a subtraction.

    • Rule: ln(ab)=ln(a)ln(b)\ln(\frac{a}{b}) = \ln(a) - \ln(b)

  • Power Rule: If the argument of the log has an exponent, that exponent can be moved to the front as a multiplier.

    • Rule: ln(xn)=nln(x)\ln(x^n) = n \ln(x)

  • Base Change Property for Aptitude Tests: Often seen in NUST and NED exams regarding terms like log10(x)\log_{10}(x).

    • The rule allows breaking a base and function into natural logs: logbase(function)=ln(function)ln(base)\log_{\text{base}}(\text{function}) = \frac{\ln(\text{function})}{\ln(\text{base})}.

    • Example: log5(6)=ln(6)ln(5)\log_5(6) = \frac{\ln(6)}{\ln(5)}. Both methods yield approximately 1.111.11.

  • Essential Constants:

    • ln(1)=0\ln(1) = 0

    • ln(e)=1\ln(e) = 1

Trigonometric Differentiation and Hashim's Memory Tricks

  • The C-Rule: All trigonometric derivatives for functions starting with the letter 'C' (Cos, Cosec, Cot) result in a negative value.

  • The Relationship Pairs: Trig functions work in couples in derivatives:

    • Sin and Cos (Couplings): ddx[sin(x)]=cos(x)\frac{d}{dx}[\sin(x)] = \cos(x) and ddx[cos(x)]=sin(x)\frac{d}{dx}[\cos(x)] = -\sin(x).

    • Tangent and Secant (The Square Relation): Tangent is linked to Secant squared. ddx[tan(x)]=sec2(x)\frac{d}{dx}[\tan(x)] = \sec^2(x).

    • Cotangent and Cosecant (The Square Relation): Cot is linked to Cosecant squared. ddx[cot(x)]=csc2(x)\frac{d}{dx}[\cot(x)] = -\csc^2(x).

    • Secant (The Product Relation): Secant differentiates into the product of itself and its pair. ddx[sec(x)]=sec(x)tan(x)\frac{d}{dx}[\sec(x)] = \sec(x) \cdot \tan(x).

    • Cosecant (The Product Relation): Cosecant differentiates into the product of itself and its pair. ddx[csc(x)]=csc(x)cot(x)\frac{d}{dx}[\csc(x)] = -\csc(x) \cdot \cot(x).

Inverse Trigonometric Differentiation

  • Rather than six separate formulas, these should be viewed as three pairs where the second function is merely the negative of the first:

    • Sin vs. Cos Inverse: ddx[sin1(x)]=11x2\frac{d}{dx}[\sin^{-1}(x)] = \frac{1}{\sqrt{1-x^2}} | ddx[cos1(x)]=11x2\frac{d}{dx}[\cos^{-1}(x)] = \frac{-1}{\sqrt{1-x^2}}

    • Tan vs. Cot Inverse: ddx[tan1(x)]=11+x2\frac{d}{dx}[\tan^{-1}(x)] = \frac{1}{1+x^2} | ddx[cot1(x)]=11+x2\frac{d}{dx}[\cot^{-1}(x)] = \frac{-1}{1+x^2}. (Note: In inverse functions, Tangent pairs with Cotangent, unlike in standard trig).

    • Sec vs. Csc Inverse: ddx[sec1(x)]=1xx21\frac{d}{dx}[\sec^{-1}(x)] = \frac{1}{x \sqrt{x^2-1}} | ddx[csc1(x)]=1xx21\frac{d}{dx}[\csc^{-1}(x)] = \frac{-1}{x \sqrt{x^2-1}}.

Hyperbolic Functions: Definitions and Derivatives

  • Hyperbolic Definitions: These functions involve natural exponentials.

    • sinh(x)=exex2\sinh(x) = \frac{e^x - e^{-x}}{2}

    • cosh(x)=ex+ex2\cosh(x) = \frac{e^x + e^{-x}}{2}

    • tanh(x)=exexex+ex\tanh(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} (Calculated by dividing Sin by Cos).

  • Derivatives of Hyperbolic Functions:

    • ddx[sinh(x)]=cosh(x)\frac{d}{dx}[\sinh(x)] = \cosh(x)

    • ddx[cosh(x)]=sinh(x)\frac{d}{dx}[\cosh(x)] = \sinh(x) (Note: This is positive, unlike the standard trigonometric derivative for Cos).

    • ddx[tanh(x)]=sech2(x)\frac{d}{dx}[\tanh(x)] = \text{sech}^2(x)

    • ddx[coth(x)]=csch2(x)\frac{d}{dx}[\coth(x)] = -\text{csch}^2(x)

    • ddx[sech(x)]=sech(x)tanh(x)\frac{d}{dx}[\text{sech}(x)] = -\text{sech}(x) \cdot \tanh(x)

    • ddx[csch(x)]=csch(x)coth(x)\frac{d}{dx}[\text{csch}(x)] = -\text{csch}(x) \cdot \coth(x)

Questions & Discussion

  • Functional Exponents (f(x)g(x)f(x)^{g(x)}): A common question involves a function as a base and another function as an exponent (e.g., cos(x)sin(x)\cos(x)^{\sin(x)}). While Intermediate textbooks (Class 12, Exercise 4.3) often require two pages of calculation, short tricks exist to solve these in one line. These will be detailed in the upcoming session on exponentials.

  • Student Inquiry: Why do 80% find this hard? The speaker reiterates that students lack a proper teacher who explains the structure of formulas, resulting in the subject feeling like an impossible "black box" of rote learning.

  • Support & Promotion: The instructor, Mohammad Hashim (Math-Netflix), emphasizes that student support is vital for new educational channels. He encourages sharing the video on WhatsApp, Facebook, and Instagram to help the Pakistani student community prepare for Round 1 of the NUST exams in December and the upcoming NED tests.