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Continuity and Differentiability - Derivative of composite functions, chain rule

Grade 12CBSE

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

🔑Concepts

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Composite Functions: A composite function is formed when one function is applied to the result of another, written as (f∘g)(x)=f(g(x))(f \circ g)(x) = f(g(x)). Visually, this can be imagined as a 'function machine' inside another machine, where the input xx passes through gg first, and the output g(x)g(x) becomes the direct input for ff.

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The Chain Rule: This rule is used to find the derivative of a composite function. It states that if yy is a function of uu, and uu is a function of xx, then the rate of change of yy with respect to xx is the product of the rate of change of yy with respect to uu and the rate of change of uu with respect to xx.

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Outside-In Differentiation: When applying the chain rule, you differentiate the 'outer' function first while keeping the 'inner' function unchanged, then multiply by the derivative of the 'inner' function. Visually, this is like peeling an onion, layer by layer, starting from the outermost shell to the core.

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Leibniz Notation: In Leibniz notation, if y=f(u)y = f(u) and u=g(x)u = g(x), the rule is expressed as dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}. This notation is helpful because it visually suggests that the dudu terms 'cancel' out, leaving dydx\frac{dy}{dx}.

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Generalized Power Rule: A specific application of the chain rule used for functions of the form y=[f(x)]ny = [f(x)]^n. The derivative is n[f(x)]n−1⋅f′(x)n[f(x)]^{n-1} \cdot f'(x). This shows how the power rule and chain rule combine to handle complex algebraic expressions.

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Repeated Chain Rule: For deeply nested functions like f(g(h(x)))f(g(h(x))), the chain rule is applied multiple times sequentially. The derivative is f′(g(h(x)))⋅g′(h(x))⋅h′(x)f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x). Conceptually, this represents a chain of dependencies where each link's change affects the next.

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Differentiability Condition: A composite function f(g(x))f(g(x)) is differentiable at a point x=cx = c if gg is differentiable at cc and ff is differentiable at g(c)g(c). If either 'link' in the chain is broken (non-differentiable), the whole composite function may fail to have a derivative at that point.

📐Formulae

dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

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

ddx[un]=nun−1⋅dudx\frac{d}{dx} [u^n] = n u^{n-1} \cdot \frac{du}{dx}

ddx[sin⁡(u)]=cos⁡(u)⋅dudx\frac{d}{dx} [\sin(u)] = \cos(u) \cdot \frac{du}{dx}

ddx[eu]=eu⋅dudx\frac{d}{dx} [e^{u}] = e^{u} \cdot \frac{du}{dx}

ddx[log⁡(u)]=1u⋅dudx\frac{d}{dx} [\log(u)] = \frac{1}{u} \cdot \frac{du}{dx}

ddx[tan⁡(u)]=sec⁡2(u)⋅dudx\frac{d}{dx} [\tan(u)] = \sec^2(u) \cdot \frac{du}{dx}

💡Examples

Problem 1:

Differentiate y=sin⁡(x2+5)y = \sin(x^2 + 5) with respect to xx.

Solution:

Step 1: Identify the inner and outer functions. Let u=x2+5u = x^2 + 5 (inner) and y=sin⁡(u)y = \sin(u) (outer). Step 2: Find the derivative of the outer function with respect to uu: dydu=cos⁡(u)\frac{dy}{du} = \cos(u). Step 3: Find the derivative of the inner function with respect to xx: dudx=ddx(x2+5)=2x\frac{du}{dx} = \frac{d}{dx}(x^2 + 5) = 2x. Step 4: Apply the chain rule: dydx=dydu⋅dudx=cos⁡(u)⋅2x\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \cos(u) \cdot 2x. Step 5: Substitute back u=x2+5u = x^2 + 5: dydx=2xcos⁡(x2+5)\frac{dy}{dx} = 2x \cos(x^2 + 5).

Explanation:

This problem uses the basic chain rule for a trigonometric composite function. We differentiate the sine function (outer) to get cosine, then multiply by the derivative of the quadratic expression (inner).

Problem 2:

Find dydx\frac{dy}{dx} if y=(3x2−7x+3)5y = (3x^2 - 7x + 3)^5.

Solution:

Step 1: Let the inner function be u=3x2−7x+3u = 3x^2 - 7x + 3. Then y=u5y = u^5. Step 2: Differentiate yy with respect to uu using the power rule: dydu=5u4\frac{dy}{du} = 5u^4. Step 3: Differentiate uu with respect to xx: dudx=6x−7\frac{du}{dx} = 6x - 7. Step 4: Combine using the chain rule: dydx=5u4⋅(6x−7)\frac{dy}{dx} = 5u^4 \cdot (6x - 7). Step 5: Substitute uu back into the equation: dydx=5(3x2−7x+3)4(6x−7)\frac{dy}{dx} = 5(3x^2 - 7x + 3)^4 (6x - 7).

Explanation:

This example demonstrates the generalized power rule. The entire polynomial is treated as a single variable uu raised to the 5th power, and the result is scaled by the derivative of that polynomial.