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Continuity and Differentiability - Differentiability

Grade 12CBSE

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

🔑Concepts

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A function f(x)f(x) is said to be differentiable at a point x=cx = c in its domain if the derivative f′(c)=lim⁡h→0f(c+h)−f(c)hf'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h} exists finitely.

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Left Hand Derivative (LHD) is defined as Lf′(c)=lim⁡h→0−f(c+h)−f(c)hL f'(c) = \lim_{h \to 0^{-}} \frac{f(c+h) - f(c)}{h} and Right Hand Derivative (RHD) as Rf′(c)=lim⁡h→0+f(c+h)−f(c)hR f'(c) = \lim_{h \to 0^{+}} \frac{f(c+h) - f(c)}{h}. For differentiability, LHD=RHDLHD = RHD.

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A fundamental theorem states: If a function is differentiable at a point, it is also continuous at that point. However, the converse is not necessarily true (e.g., f(x)=∣x∣f(x) = |x| is continuous at x=0x=0 but not differentiable).

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The Chain Rule is used for composite functions: If y=f(u)y = f(u) and u=g(x)u = g(x), then dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}.

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Implicit Differentiation involves differentiating both sides of an equation with respect to xx when yy cannot be easily expressed as an explicit function of xx.

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Logarithmic Differentiation is useful for functions of the form y=[f(x)]g(x)y = [f(x)]^{g(x)} or products of multiple functions. We take log⁡\log on both sides before differentiating.

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Parametric Differentiation: If x=f(t)x = f(t) and y=g(t)y = g(t), then dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}, provided dxdt≠0\frac{dx}{dt} \neq 0.

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Second Order Derivative: If y=f(x)y = f(x), then d2ydx2=ddx(dydx)\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right).

📐Formulae

f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

ddx(uv)=udvdx+vdudx\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}

ddx(uv)=vdudx−udvdxv2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}

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

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

ddx(ax)=axlog⁡ea\frac{d}{dx}(a^x) = a^x \log_e a

ddx(log⁡ex)=1x\frac{d}{dx}(\log_e x) = \frac{1}{x}

💡Examples

Problem 1:

Examine the differentiability of f(x)=∣x−1∣f(x) = |x - 1| at x=1x = 1.

Solution:

We check LHDLHD and RHDRHD at x=1x = 1. Lf′(1)=lim⁡h→0−∣(1+h)−1∣−∣1−1∣h=lim⁡h→0−∣h∣hL f'(1) = \lim_{h \to 0^{-}} \frac{|(1+h)-1| - |1-1|}{h} = \lim_{h \to 0^{-}} \frac{|h|}{h}. Since h<0h < 0, ∣h∣=−h|h| = -h. So, LHD=lim⁡h→0−−hh=−1LHD = \lim_{h \to 0^{-}} \frac{-h}{h} = -1. Rf′(1)=lim⁡h→0+∣(1+h)−1∣−∣1−1∣h=lim⁡h→0+∣h∣hR f'(1) = \lim_{h \to 0^{+}} \frac{|(1+h)-1| - |1-1|}{h} = \lim_{h \to 0^{+}} \frac{|h|}{h}. Since h>0h > 0, ∣h∣=h|h| = h. So, RHD=lim⁡h→0+hh=1RHD = \lim_{h \to 0^{+}} \frac{h}{h} = 1. Since LHD≠RHDLHD \neq RHD, f(x)f(x) is not differentiable at x=1x = 1.

Explanation:

Differentiability requires the slope from the left and right to be equal. Here, the 'V' shape of the absolute value function creates a sharp corner at x=1x=1 where slopes differ.

Problem 2:

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

Solution:

Differentiate xx with respect to θ\theta: dxdθ=a(1+cos⁡θ)\frac{dx}{d\theta} = a(1 + \cos \theta). Differentiate yy with respect to θ\theta: dydθ=a(0−(−sin⁡θ))=asin⁡θ\frac{dy}{d\theta} = a(0 - (-\sin \theta)) = a \sin \theta. Now, dydx=dy/dθdx/dθ=asin⁡θa(1+cos⁡θ)\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a \sin \theta}{a(1 + \cos \theta)}. Using identities sin⁡θ=2sin⁡θ2cos⁡θ2\sin \theta = 2 \sin\frac{\theta}{2} \cos\frac{\theta}{2} and 1+cos⁡θ=2cos⁡2θ21 + \cos \theta = 2 \cos^2\frac{\theta}{2}: dydx=2sin⁡θ2cos⁡θ22cos⁡2θ2=tan⁡θ2\frac{dy}{dx} = \frac{2 \sin\frac{\theta}{2} \cos\frac{\theta}{2}}{2 \cos^2\frac{\theta}{2}} = \tan\frac{\theta}{2}.

Explanation:

This uses parametric differentiation and trigonometric simplification to find the derivative of yy with respect to xx.

Problem 3:

If y=xxy = x^x, find dydx\frac{dy}{dx}.

Solution:

Taking natural logarithm on both sides: log⁡y=log⁡(xx)  ⟹  log⁡y=xlog⁡x\log y = \log(x^x) \implies \log y = x \log x. Differentiating both sides with respect to xx: 1ydydx=ddx(xlog⁡x)\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x \log x). Applying product rule: 1ydydx=x⋅1x+log⁡x⋅1=1+log⁡x\frac{1}{y} \frac{dy}{dx} = x \cdot \frac{1}{x} + \log x \cdot 1 = 1 + \log x. Therefore, dydx=y(1+log⁡x)=xx(1+log⁡x)\frac{dy}{dx} = y(1 + \log x) = x^x(1 + \log x).

Explanation:

Logarithmic differentiation is required here because the base and the exponent are both variables.