krit.club logo

Calculus - Logarithmic and Parametric Differentiation

Grade 12ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Logarithmic Differentiation is a technique used to differentiate functions where the variable appears in the exponent, such as y=[f(x)]g(x)y = [f(x)]^{g(x)}, or for functions consisting of products and quotients of several terms. Visually, taking the logarithm 'flattens' the powers and converts products into sums, making the function easier to differentiate.

•

The core process of Logarithmic Differentiation involves three steps: first, taking the natural logarithm (ln⁡\ln) on both sides; second, using log properties to simplify the expression; and third, differentiating both sides with respect to xx using implicit differentiation and the Chain Rule.

•

Properties of Logarithms are the foundation of this method. These include the product rule ln⁡(mn)=ln⁡m+ln⁡n\ln(mn) = \ln m + \ln n, the quotient rule ln⁡(m/n)=ln⁡m−ln⁡n\ln(m/n) = \ln m - \ln n, and the power rule ln⁡(mn)=nln⁡m\ln(m^n) = n \ln m. These properties allow us to break down complex algebraic structures into linear components.

•

Parametric Differentiation is used when xx and yy are both expressed as functions of a third variable, called a parameter (usually tt or θ\theta). For example, x=f(t)x = f(t) and y=g(t)y = g(t). Visually, this describes a curve in the xyxy-plane where the position depends on a parameter like time.

•

To find the slope of the tangent (dydx\frac{dy}{dx}) for parametric equations, we use the ratio of their derivatives with respect to the parameter: dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}. Geometrically, this represents the ratio of vertical 'velocity' to horizontal 'velocity' at any given point on the curve.

•

Second-order derivatives in parametric form require extra care. To find d2ydx2\frac{d^2y}{dx^2}, you must differentiate the first derivative (dydx\frac{dy}{dx}) with respect to the parameter tt, and then divide the result by dxdt\frac{dx}{dt}. A common visual error is to simply take the second derivative of yy over the second derivative of xx, which is mathematically incorrect.

•

Implicit Differentiation is often the final step in logarithmic differentiation. Since we start with ln⁡y\ln y, the derivative of the left side is always 1y⋅dydx\frac{1}{y} \cdot \frac{dy}{dx} because yy is a function of xx. This represents the rate of change of the log-transformed variable relative to the original function's value.

📐Formulae

ln⁡(ab)=ln⁡a+ln⁡v\ln(ab) = \ln a + \ln v

ln⁡(ab)=ln⁡a−ln⁡b\ln(\frac{a}{b}) = \ln a - \ln b

ln⁡(ab)=bln⁡a\ln(a^b) = b \ln a

ddx(ln⁡y)=1y⋅dydx\frac{d}{dx}(\ln y) = \frac{1}{y} \cdot \frac{dy}{dx}

ddx(xn)=nxn−1\frac{d}{dx}(x^n) = nx^{n-1}

dydx=dydtdxdt, where dxdt≠0\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}, \text{ where } \frac{dx}{dt} \neq 0

d2ydx2=ddt(dydx)⋅dtdx=ddt(dydx)dxdt\frac{d^2y}{dx^2} = \frac{d}{dt} (\frac{dy}{dx}) \cdot \frac{dt}{dx} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}

💡Examples

Problem 1:

Differentiate y=xsin⁡xy = x^{\sin x} with respect to xx.

Solution:

Step 1: Take natural log on both sides: ln⁡y=ln⁡(xsin⁡x)\ln y = \ln(x^{\sin x}) Step 2: Use the log power property: ln⁡y=sin⁡x⋅ln⁡x\ln y = \sin x \cdot \ln x Step 3: Differentiate both sides with respect to xx using the Product Rule on the right: 1ydydx=sin⁡x⋅ddx(ln⁡x)+ln⁡x⋅ddx(sin⁡x)\frac{1}{y} \frac{dy}{dx} = \sin x \cdot \frac{d}{dx}(\ln x) + \ln x \cdot \frac{d}{dx}(\sin x) 1ydydx=sin⁡x⋅1x+ln⁡x⋅cos⁡x\frac{1}{y} \frac{dy}{dx} = \sin x \cdot \frac{1}{x} + \ln x \cdot \cos x Step 4: Multiply by yy to isolate dydx\frac{dy}{dx}: dydx=y[sin⁡xx+cos⁡xln⁡x]\frac{dy}{dx} = y [\frac{\sin x}{x} + \cos x \ln x] Step 5: Substitute the original expression for yy: dydx=xsin⁡x(sin⁡xx+cos⁡xln⁡x)\frac{dy}{dx} = x^{\sin x} (\frac{\sin x}{x} + \cos x \ln x)

Explanation:

Since the function has a variable in both the base and the exponent, we apply logarithmic differentiation. We use the property ln⁡(ab)=bln⁡a\ln(a^b) = b \ln a to bring the exponent down and then apply the product rule.

Problem 2:

Find dydx\frac{dy}{dx} if x=a(t+sin⁡t)x = a(t + \sin t) and y=a(1−cos⁡t)y = a(1 - \cos t).

Solution:

Step 1: Differentiate xx with respect to tt: dxdt=a(1+cos⁡t)\frac{dx}{dt} = a(1 + \cos t) Step 2: Differentiate yy with respect to tt: dydt=a(0−(−sin⁡t))=asin⁡t\frac{dy}{dt} = a(0 - (-\sin t)) = a \sin t Step 3: Apply the parametric derivative formula: dydx=dy/dtdx/dt=asin⁡ta(1+cos⁡t)\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a \sin t}{a(1 + \cos t)} Step 4: Simplify using trigonometric identities (sin⁡t=2sin⁡t2cos⁡t2\sin t = 2 \sin \frac{t}{2} \cos \frac{t}{2} and 1+cos⁡t=2cos⁡2t21 + \cos t = 2 \cos^2 \frac{t}{2}): dydx=2sin⁡t2cos⁡t22cos⁡2t2\frac{dy}{dx} = \frac{2 \sin \frac{t}{2} \cos \frac{t}{2}}{2 \cos^2 \frac{t}{2}} dydx=tan⁡t2\frac{dy}{dx} = \tan \frac{t}{2}

Explanation:

The equations define a cycloid. We find the derivatives of xx and yy independently with respect to the parameter tt, then divide them. Trigonometric simplification is used to reach the final elegant form.