Comprehensive Introduction to Calculus and Differentiation Rules

The Fundamental Power Rule of Differentiation

  • The primary rule of differentiation discussed is the Power Rule. If a function is defined as y=xny = x^n, its derivative with respect to xx is given by:     dydx=nxn−1\frac{dy}{dx} = n x^{n-1}

  • The process involves taking the original exponent (nn), multiplying the base (xx) by that exponent, and then subtracting one (11) from the original exponent to find the new power.

  • This rule applies to positive, negative, and fractional exponents.

Calcu the power rule often requires subtracting $1$ from a fraction. The transcript details several arithmetic steps for this process:lations Involving Fractional and Negative Exponents

  • Applying

    • Example 1: Subtracting one from 103\frac{10}{3}

      • 103−1=10−(3×1)3=10−33=73\frac{10}{3} - 1 = \frac{10 - (3 \times 1)}{3} = \frac{10 - 3}{3} = \frac{7}{3}

      • (Note: The transcript mentions a final result of 78\frac{7}{8}, though the arithmetic calculation shown is 103−1=73\frac{10}{3} - 1 = \frac{7}{3}.)

    • Example 2: Subtracting one from 47\frac{4}{7}

      • 47−1=4−(7×1)7=4−77=−37\frac{4}{7} - 1 = \frac{4 - (7 \times 1)}{7} = \frac{4 - 7}{7} = -\frac{3}{7}

    • Example 3: Differentiation involving a negative power −5-5

      • If y=x−5y = x^{-5}, then dydx=−5x−5−1=−5x−6\frac{dy}{dx} = -5x^{-5 - 1} = -5x^{-6}.

  • The transcript also addresses functions where the variable is in the denominator, such as y=1x2/3y = \frac{1}{x^{2/3}}. Depending on the requirements of the question, this can be rewritten as y=x−2/3y = x^{-2/3} before applying the power rule:

    • dydx=−23x−2/3−1=−23x−5/3\frac{dy}{dx} = -\frac{2}{3}x^{-2/3 - 1} = -\frac{2}{3}x^{-5/3}.

  • Further fractional subtraction examples include:

    • −45−1=−95-\frac{4}{5} - 1 = -\frac{9}{5}

    • −12−1=−32-\frac{1}{2} - 1 = -\frac{3}{2}

Differentiation of Constants

  • The symbol kk is used to represent a constant.

  • The rule for a constant is that the derivative of any constant value is always zero.

  • Rule: If y=ky = k, then dydx=0\frac{dy}{dx} = 0.

  • Examples provided in the transcript include:

    • If y=5y = 5, dydx=0\frac{dy}{dx} = 0.

    • If y=πy = \pi, dydx=0\frac{dy}{dx} = 0.

    • The constant π\pi is treated like any other numerical value in this context.

Differentiation with Coefficients and specific Powers

  • When a term has a coefficient (e.g., y=7xy = 7x), the rule is applied as follows:

    • y=7x1y = 7x^1

    • dydx=7×1x1−1=7x0\frac{dy}{dx} = 7 \times 1x^{1-1} = 7x^0

    • Since x0=1x^0 = 1, the final answer is 7×1=77 \times 1 = 7.

  • Complex Example 1: y=15x−3/5y = 15x^{-3/5}

    • Differentiation: dydx=15×(−35)x−3/5−1\frac{dy}{dx} = 15 \times (-\frac{3}{5})x^{-3/5 - 1}

    • The transcript notes that five divides into fifteen three times: 3×(−3)=−93 \times (-3) = -9.

    • The new power is −3/5−1=−8/5-3/5 - 1 = -8/5.

    • Result: −9x−8/5-9x^{-8/5}.

  • Complex Example 2: y=−94x2/3y = -\frac{9}{4}x^{2/3}

    • Differentiation involves multiplying the coefficient −9/4-9/4 by the power 2/32/3 and subtracting one from the power.

The Sum and Difference Rule

  • The derivative of a sum or difference of functions is the sum or difference of their individual derivatives.

  • General Formula: If y=u±v±wy = u \pm v \pm w, where uu, vv, and ww are functions of xx, then:     dydx=dudx±dvdx±dwdx\frac{dy}{dx} = \frac{du}{dx} \pm \frac{dv}{dx} \pm \frac{dw}{dx}

  • Example Problem: Differentiating the expression y=x3−x2+x−8y = x^3 - x^2 + x - 8

    • Step 1: Identify terms individually (x3x^3, −x2-x^2, xx, −8-8).

    • Step 2: Differentiate each term:

      • ddx(x3)=3x2\frac{d}{dx}(x^3) = 3x^2

      • ddx(x2)=2x\frac{d}{dx}(x^2) = 2x

      • ddx(x)=1\frac{d}{dx}(x) = 1

      • ddx(8)=0\frac{d}{dx}(8) = 0

    • Step 3: Combine the results: dydx=3x2−2x+1\frac{dy}{dx} = 3x^2 - 2x + 1.

  • Additional Case: y=5x2−3x+17y = 5x^2 - 3x + 17

    • dydx=(5×2)x2−1−(3×1)x1−1+0\frac{dy}{dx} = (5 \times 2)x^{2-1} - (3 \times 1)x^{1-1} + 0

    • dydx=10x−3\frac{dy}{dx} = 10x - 3.

Introduction to the Product Rule

  • The transcript introduces the basic structure for differentiating the product of two functions, uu and vv.

  • General Product Rule Formula: If y=uvy = uv, then:     dydx=udvdx+vdudx\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}

  • In this formula:

    • uu is the first function.

    • vv is the second function.

    • du/dxdu/dx and dv/dxdv/dx are the respective derivatives of those functions.

  • Example walkthrough: y=(x2)(9x−1)y = (x^2)(9x - 1)

    • Assign u=x2u = x^2 and v=9x−1v = 9x - 1.

    • Apply the formula: (x2)ddx(9x−1)+(9x−1)ddx(x2)(x^2) \frac{d}{dx}(9x - 1) + (9x - 1) \frac{d}{dx}(x^2).

    • Other values relevant to complex composite differentiation mentioned: x+12x + 12, −2/3-2/3, and −1/2-1/2.