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Calculus - Derivatives of Composite, Implicit, Inverse Trigonometric, Exponential and Logarithmic Functions

Grade 12ICSE

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

🔑Concepts

Chain Rule for Composite Functions: When a function is nested within another, such as y=f(g(x))y = f(g(x)), the derivative is calculated as the derivative of the outer function multiplied by the derivative of the inner function, expressed as dydx=f(g(x))g(x)\frac{dy}{dx} = f'(g(x)) \cdot g'(x). Visually, this process is like peeling an onion; you differentiate the outermost layer first and work your way inward. On a graph, the steepness of the composite curve is determined by the compounding rates of change of its individual components.

Implicit Differentiation: This technique is used for equations where yy is not explicitly isolated, such as x2+y2=r2x^2 + y^2 = r^2. We differentiate every term with respect to xx, and whenever we differentiate a term involving yy, we multiply by dydx\frac{dy}{dx} as per the chain rule. Geometrically, implicit differentiation allows us to find the slope of the tangent line to curves that may not be functions (like circles or ellipses) at any point (x,y)(x, y) on the curve.

Derivatives of Exponential Functions: The natural exponential function exe^x is unique because its slope at any point is exactly equal to its y-value. Visually, the graph of y=exy = e^x becomes steeper as it moves to the right, and the rate of this steepening is the function itself. For a general base aa, the derivative of axa^x is axlnaa^x \ln a, where the lna\ln a factor acts as a scaling constant for the growth rate.

Logarithmic Differentiation: This is a powerful method used for complex products, quotients, or functions of the form y=[f(x)]g(x)y = [f(x)]^{g(x)}. By taking the natural logarithm (ln)(\ln) of both sides, we use log properties to turn powers into products and products into sums. After differentiating implicitly, we solve for dydx\frac{dy}{dx}. This method linearizes the relationship between the base and the exponent, making the differentiation of variable-power functions manageable.

Derivatives of Inverse Trigonometric Functions: Functions like sin1x\sin^{-1} x and tan1x\tan^{-1} x have derivatives that are algebraic fractions. For example, the derivative of tan1x\tan^{-1} x is 11+x2\frac{1}{1+x^2}. Visually, the inverse tangent function has horizontal asymptotes at y=±π2y = \pm \frac{\pi}{2} and is strictly increasing; its derivative represents a bell-shaped curve that peaks at x=0x=0, reflecting the steepest part of the original function's graph.

Logarithmic Functions: The derivative of lnx\ln x is 1x\frac{1}{x} for x>0x > 0. Visually, the natural log curve grows slower and slower as xx increases, which is reflected in its derivative 1x\frac{1}{x} approaching zero. If the base of the log is not ee, we use the change of base formula, resulting in ddx(logax)=1xlna\frac{d}{dx}(\log_a x) = \frac{1}{x \ln a}.

📐Formulae

ddx[f(g(x))]=f(g(x))g(x)\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)

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

ddx(ax)=axlna\frac{d}{dx}(a^x) = a^x \ln a

ddx(lnx)=1x\frac{d}{dx}(\ln x) = \frac{1}{x}

ddx(logax)=1xlna\frac{d}{dx}(\log_a x) = \frac{1}{x \ln a}

ddx(sin1x)=11x2\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}}

ddx(cos1x)=11x2\frac{d}{dx}(\cos^{-1} x) = -\frac{1}{\sqrt{1-x^2}}

ddx(tan1x)=11+x2\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}

ddx(cot1x)=11+x2\frac{d}{dx}(\cot^{-1} x) = -\frac{1}{1+x^2}

ddx(sec1x)=1xx21\frac{d}{dx}(\sec^{-1} x) = \frac{1}{|x|\sqrt{x^2-1}}

💡Examples

Problem 1:

Find dydx\frac{dy}{dx} if y=(sinx)cosxy = (\sin x)^{\cos x}.

Solution:

  1. Take the natural log of both sides: lny=ln((sinx)cosx)\ln y = \ln ((\sin x)^{\cos x})
  2. Use log properties: lny=cosxln(sinx)\ln y = \cos x \cdot \ln(\sin x)
  3. Differentiate both sides with respect to xx using the Product Rule on the right: 1ydydx=ddx(cosx)ln(sinx)+cosxddx(ln(sinx))\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\cos x) \cdot \ln(\sin x) + \cos x \cdot \frac{d}{dx}(\ln(\sin x))
  4. Compute derivatives: 1ydydx=sinxln(sinx)+cosx1sinxcosx\frac{1}{y} \frac{dy}{dx} = -\sin x \ln(\sin x) + \cos x \cdot \frac{1}{\sin x} \cdot \cos x
  5. Simplify: 1ydydx=sinxln(sinx)+cosxcotx\frac{1}{y} \frac{dy}{dx} = -\sin x \ln(\sin x) + \cos x \cot x
  6. Multiply by yy: dydx=y[cosxcotxsinxln(sinx)]\frac{dy}{dx} = y [\cos x \cot x - \sin x \ln(\sin x)]
  7. Substitute original yy: dydx=(sinx)cosx[cosxcotxsinxln(sinx)]\frac{dy}{dx} = (\sin x)^{\cos x} [\cos x \cot x - \sin x \ln(\sin x)]

Explanation:

This problem uses logarithmic differentiation because the function has a variable in both the base and the exponent. Taking ln\ln converts the exponentiation into a product, which is then solvable via the Product Rule and Chain Rule.

Problem 2:

Find dydx\frac{dy}{dx} for the implicit equation x2+xy+y2=10x^2 + xy + y^2 = 10.

Solution:

  1. Differentiate both sides with respect to xx: ddx(x2)+ddx(xy)+ddx(y2)=ddx(10)\frac{d}{dx}(x^2) + \frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(10)
  2. Apply the Product Rule to xyxy and Chain Rule to y2y^2: 2x+(xdydx+y1)+2ydydx=02x + (x \frac{dy}{dx} + y \cdot 1) + 2y \frac{dy}{dx} = 0
  3. Group terms containing dydx\frac{dy}{dx}: xdydx+2ydydx=2xyx \frac{dy}{dx} + 2y \frac{dy}{dx} = -2x - y
  4. Factor out dydx\frac{dy}{dx}: dydx(x+2y)=(2x+y)\frac{dy}{dx}(x + 2y) = -(2x + y)
  5. Solve for dydx\frac{dy}{dx}: dydx=2x+yx+2y\frac{dy}{dx} = -\frac{2x + y}{x + 2y}

Explanation:

This example demonstrates implicit differentiation. Since yy cannot be easily isolated, we differentiate term-by-term. The term xyxy requires the product rule, and y2y^2 requires the chain rule (multiplying by dydx\frac{dy}{dx}).