Differentiability and Chain Rule

Differentiability

Let $f$ be a real-valued function and let $c$ be a point in its domain. The derivative of $f$ at $c$ is defined by$$f'(c) = \lim_{h \to 0} \frac{f(c+h)-f(c)}{h},$$ provided this limit exists as a finite real number.

The derivative of $f$ at $c$ is denoted by $f'(c)$ or $$\left.\frac{d}{dx}[f(x)]\right|_{x=c}.$$

For a function to be differentiable at a point $c$, both the left-hand derivative and the right-hand derivative must exist and be equal.

• Left-Hand Derivative (LHD): $$\lim_{h \to 0^-} \frac{f(c+h)-f(c)}{h}$$
• Right-Hand Derivative (RHD): $$\lim_{h \to 0^+} \frac{f(c+h)-f(c)}{h}$$

If LHD = RHD, then the common value is the derivative $f'(c)$.

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Relationship between Continuity and Differentiability

Theorem 3: If a function $f$ is differentiable at a point $c$, then it is also continuous at that point.

Reason: If $$f'(c)=\lim_{h\to 0}\frac{f(c+h)-f(c)}{h}$$ exists, then the quotient remains finite as $h \to 0$. Since $h \to 0$, the numerator $f(c+h)-f(c)$ must also tend to $0$. Therefore, $$\lim_{h\to 0} f(c+h)=f(c),$$ which means $f$ is continuous at $c$.

Corollary: Every differentiable function is continuous.

Important Note: The converse is not true. A function may be continuous at a point but not differentiable there. A standard example is $$f(x)=|x|,$$ which is continuous at $x=0$ but not differentiable at $x=0$ because the graph has a sharp corner there.

Thus, $$\text{Differentiability} \implies \text{Continuity},$$ but $$\text{Continuity} \not\implies \text{Differentiability}.$$

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Algebra of Derivatives

Let $u$ and $v$ be differentiable functions of $x$. Then the standard rules of differentiation are:

1. Sum/Difference Rule: $$(u \pm v)' = u' \pm v'$$

2. Product Rule (Leibnitz Rule): $$(uv)' = u'v + uv'$$

3. Quotient Rule: $$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}, \quad v \ne 0.$$

These rules are used repeatedly along with the chain rule when differentiating composite expressions.

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Chain Rule

The Chain Rule is used to differentiate a composite function.

Theorem 4 (Chain Rule): Let a function be written in the form $$y = f(x) = v(u(x)).$$ If we put $$t = u(x),$$ then $$y = v(t).$$ If both $\dfrac{dt}{dx}$ and $\dfrac{dy}{dt}$ exist, then $$\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}.$$

In Leibniz notation, if $y$ depends on $t$ and $t$ depends on $x$, then the derivative of $y$ with respect to $x$ is the product of the derivative of $y$ with respect to $t$ and the derivative of $t$ with respect to $x$.

For a composition of three or more functions, the chain rule is applied step by step. For example, if $$y = w(u(v(x))),$$ and we let $$t = v(x), \qquad s = u(t),$$ then $$\frac{dy}{dx} = \frac{dy}{ds} \cdot \frac{ds}{dt} \cdot \frac{dt}{dx}.$$

Solved Examples

Example 1: Non-differentiability of Modulus Function

Prove that the function $f$ given by $f(x) = |x-1|, x \in \mathbb{R}$ is not differentiable at $x=1$.

Solution: To test differentiability at $x=1$, we compare the left-hand derivative and right-hand derivative.

Step 1: Find the function value at $x=1$: $$f(1)=|1-1|=0.$$

Step 2: Compute the left-hand derivative: $$LHD = \lim_{h \to 0^-} \frac{f(1+h)-f(1)}{h}.$$ Now, $$f(1+h)=|(1+h)-1|=|h|.$$ For $h<0$, we have $|h|=-h$. Therefore, $$LHD = \lim_{h \to 0^-} \frac{-h-0}{h} = \lim_{h \to 0^-} (-1) = -1.$$

Step 3: Compute the right-hand derivative: $$RHD = \lim_{h \to 0^+} \frac{f(1+h)-f(1)}{h}.$$ For $h>0$, we have $|h|=h$. Hence, $$RHD = \lim_{h \to 0^+} \frac{h-0}{h} = \lim_{h \to 0^+} 1 = 1.$$

Step 4: Compare LHD and RHD: $$LHD = -1, \qquad RHD = 1.$$ Since the left-hand and right-hand derivatives are not equal, the derivative does not exist at $x=1$.

Therefore, the function $f(x)=|x-1|$ is not differentiable at $x=1$.

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Example 2: Basic Chain Rule

Find the derivative of the function given by $f(x) = \sin(x^2)$.

Solution: This is a composite function. Let $$t = x^2.$$ Then $$y = \sin t.$$

Step 1: Differentiate $t$ with respect to $x$: $$\frac{dt}{dx} = 2x.$$

Step 2: Differentiate $y$ with respect to $t$: $$\frac{dy}{dt} = \cos t.$$

Step 3: Apply the chain rule: $$\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = \cos t \cdot 2x.$$

Step 4: Substitute $t=x^2$ back: $$\frac{dy}{dx} = 2x\cos(x^2).$$

Answer: $$2x\cos(x^2)$$

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Example 3: Chain Rule with Linear Argument

Find the derivative of $y = \tan(2x+3)$.

Solution: Let $$t = 2x+3.$$ Then $$y = \tan t.$$

Step 1: Differentiate $t$ with respect to $x$: $$\frac{dt}{dx}=2.$$

Step 2: Differentiate $y$ with respect to $t$: $$\frac{dy}{dt}=\sec^2 t.$$

Step 3: Apply the chain rule: $$\frac{dy}{dx}=\frac{dy}{dt}\cdot\frac{dt}{dx} = \sec^2 t \cdot 2.$$

Step 4: Replace $t$ by $2x+3$: $$\frac{dy}{dx}=2\sec^2(2x+3).$$

Answer: $$2\sec^2(2x+3)$$

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Example 4: Composition of Three Functions

Differentiate $\sin(\cos(x^2))$ with respect to $x$.

Solution: Let $$y = \sin(\cos(x^2)).$$ This is a composition of three functions:

1. inner function: $x^2$,
2. middle function: $\cos(\cdot)$,
3. outer function: $\sin(\cdot)$.

Step 1: Differentiate the outermost sine function: $$\frac{dy}{dx} = \cos(\cos(x^2)) \cdot \frac{d}{dx}[\cos(x^2)].$$

Step 2: Differentiate the middle cosine function: $$\frac{d}{dx}[\cos(x^2)] = -\sin(x^2) \cdot \frac{d}{dx}(x^2).$$

Step 3: Differentiate the inner function: $$\frac{d}{dx}(x^2)=2x.$$

Step 4: Combine all factors: $$\frac{dy}{dx} = \cos(\cos(x^2)) \cdot (-\sin(x^2)) \cdot 2x.$$ So, $$\frac{dy}{dx} = -2x\sin(x^2)\cos(\cos(x^2)).$$

Answer: $$-2x\sin(x^2)\cos(\cos(x^2))$$

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Example 5: Product Rule combined with Chain Rule

Differentiate $y = \sin^3 x \cdot \cos^6 x$ with respect to x.

Solution: Let $$u = \sin^3 x, \qquad v = \cos^6 x.$$ Then $$y = uv.$$

Step 1: Apply the product rule: $$\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}.$$ So, $$\frac{dy}{dx} = \sin^3 x \cdot \frac{d}{dx}(\cos^6 x) + \cos^6 x \cdot \frac{d}{dx}(\sin^3 x).$$

Step 2: Differentiate $\cos^6 x$ using the chain rule: $$\frac{d}{dx}(\cos^6 x) = 6\cos^5 x \cdot \frac{d}{dx}(\cos x) = 6\cos^5 x(-\sin x) = -6\sin x\cos^5 x.$$

Step 3: Differentiate $\sin^3 x$ using the chain rule: $$\frac{d}{dx}(\sin^3 x) = 3\sin^2 x \cdot \frac{d}{dx}(\sin x) = 3\sin^2 x\cos x.$$

Step 4: Substitute back: $$\frac{dy}{dx} = \sin^3 x(-6\sin x\cos^5 x) + \cos^6 x(3\sin^2 x\cos x).$$ That is, $$\frac{dy}{dx} = -6\sin^4 x\cos^5 x + 3\sin^2 x\cos^7 x.$$

Step 5: Factor out the common term $3\sin^2 x\cos^5 x$: $$\frac{dy}{dx} = 3\sin^2 x\cos^5 x(-2\sin^2 x + \cos^2 x).$$ So we may write $$\frac{dy}{dx} = 3\sin^2 x\cos^5 x(\cos^2 x - 2\sin^2 x).$$

Answer: $$3\sin^2 x\cos^5 x(\cos^2 x - 2\sin^2 x)$$

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Example 6: Chain Rule with Roots

Find the derivative of $y = 2\sqrt{\cot(x^2)}$.

Solution: First rewrite the square root as a power: $$y = 2[\cot(x^2)]^{1/2}.$$

Step 1: Differentiate using the power rule and chain rule: $$\frac{dy}{dx} = 2 \cdot \frac{1}{2}[\cot(x^2)]^{-1/2} \cdot \frac{d}{dx}[\cot(x^2)].$$ So, $$\frac{dy}{dx} = [\cot(x^2)]^{-1/2} \cdot \frac{d}{dx}[\cot(x^2)].$$ That is, $$\frac{dy}{dx} = \frac{1}{\sqrt{\cot(x^2)}} \cdot \frac{d}{dx}[\cot(x^2)].$$

Step 2: Differentiate $\cot(x^2)$: $$\frac{d}{dx}[\cot(x^2)] = -\csc^2(x^2) \cdot \frac{d}{dx}(x^2).$$ Since $$\frac{d}{dx}(x^2)=2x,$$ we get $$\frac{d}{dx}[\cot(x^2)] = -2x\csc^2(x^2).$$

Step 3: Substitute back: $$\frac{dy}{dx} = \frac{-2x\csc^2(x^2)}{\sqrt{\cot(x^2)}}.$$

Answer: $$\frac{-2x\csc^2(x^2)}{\sqrt{\cot(x^2)}}$$

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Example 7: Differentiating Composite Roots

Differentiate $\cos(\sqrt{x})$ with respect to $x$.

Solution: Let $$y = \cos(\sqrt{x}).$$

Step 1: Apply the chain rule to the outer cosine function: $$\frac{dy}{dx} = -\sin(\sqrt{x}) \cdot \frac{d}{dx}(\sqrt{x}).$$

Step 2: Differentiate $\sqrt{x}$: $$\sqrt{x} = x^{1/2}, \qquad \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}.$$

Step 3: Substitute back: $$\frac{dy}{dx} = -\sin(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}}.$$ Hence, $$\frac{dy}{dx} = \frac{-\sin(\sqrt{x})}{2\sqrt{x}}.$$

Answer: $$\frac{-\sin(\sqrt{x})}{2\sqrt{x}}$$

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Example 8: Non-differentiability of Greatest Integer Function

Prove that the greatest integer function defined by $f(x) = [x], 0 < x < 3$ is not differentiable at $x=1$ and $x=2$.

Solution: A function must first be continuous in order to be differentiable. The greatest integer function has jump discontinuities at integers, so it cannot be differentiable there. We can also verify this directly using one-sided derivatives.

At $x=1$

Step 1: Note that $$f(1) = [1] = 1.$$ For $h<0$ and close to 0, $1+h<1$, so $$[1+h]=0.$$ Thus, $$LHD = \lim_{h \to 0^-} \frac{f(1+h)-f(1)}{h} = \lim_{h \to 0^-} \frac{0-1}{h} = \lim_{h \to 0^-} \frac{-1}{h}.$$ As $h \to 0^-$, the quantity $\frac{-1}{h} \to +\infty$, so the left-hand derivative does not exist as a finite number.

For $h>0$ and close to 0, $1+h>1$ but less than 2, so $$[1+h]=1.$$ Hence, $$RHD = \lim_{h \to 0^+} \frac{1-1}{h} = 0.$$ Since LHD and RHD are not equal, the function is not differentiable at $x=1$.

At $x=2$

Step 2: Note that $$f(2) = [2] = 2.$$ For $h<0$ and close to 0, $2+h<2$ but greater than 1, so $$[2+h]=1.$$ Then, $$LHD = \lim_{h \to 0^-} \frac{1-2}{h} = \lim_{h \to 0^-} \frac{-1}{h} \to +\infty,$$ so the left-hand derivative does not exist as a finite number.

For $h>0$ and close to 0, $2+h>2$ but less than 3, so $$[2+h]=2.$$ Thus, $$RHD = \lim_{h \to 0^+} \frac{2-2}{h} = 0.$$ Again LHD and RHD are not equal, so the function is not differentiable at $x=2$.

Therefore, $f(x)=[x]$ is not differentiable at both $x=1$ and $x=2$.

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Example 9: Differentiating Secant and Tangent Composition

Differentiate $\sec(\tan(\sqrt{x}))$ with respect to $x$.

Solution: Let $$y = \sec(\tan(\sqrt{x})).$$ This is a composition of three functions: $\sqrt{x}$, then $\tan$, then $\sec$.

Step 1: Differentiate the outermost secant function: $$\frac{dy}{dx} = \sec(\tan(\sqrt{x}))\tan(\tan(\sqrt{x})) \cdot \frac{d}{dx}[\tan(\sqrt{x})].$$

Step 2: Differentiate $\tan(\sqrt{x})$: $$\frac{d}{dx}[\tan(\sqrt{x})] = \sec^2(\sqrt{x}) \cdot \frac{d}{dx}(\sqrt{x}).$$

Step 3: Differentiate $\sqrt{x}$: $$\frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}.$$

Step 4: Combine all parts: $$\frac{dy}{dx} = \sec(\tan(\sqrt{x}))\tan(\tan(\sqrt{x}))\sec^2(\sqrt{x})\cdot \frac{1}{2\sqrt{x}}.$$ Therefore, $$\frac{dy}{dx} = \frac{\sec(\tan(\sqrt{x}))\tan(\tan(\sqrt{x}))\sec^2(\sqrt{x})}{2\sqrt{x}}.$$

Answer: $$\frac{\sec(\tan(\sqrt{x}))\tan(\tan(\sqrt{x}))\sec^2(\sqrt{x})}{2\sqrt{x}}$$

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Example 10: Quotient Rule with Chain Rule

Differentiate $y = \frac{\sin(ax+b)}{\cos(cx+d)}$.

Solution: Let $$u = \sin(ax+b), \qquad v = \cos(cx+d).$$ Then $$y = \frac{u}{v}.$$

Step 1: Apply the quotient rule: $$\frac{dy}{dx} = \frac{u'v - uv'}{v^2}.$$

Step 2: Differentiate $u$ using the chain rule: $$u' = \frac{d}{dx}[\sin(ax+b)] = \cos(ax+b) \cdot a = a\cos(ax+b).$$

Step 3: Differentiate $v$ using the chain rule: $$v' = \frac{d}{dx}[\cos(cx+d)] = -\sin(cx+d) \cdot c = -c\sin(cx+d).$$

Step 4: Substitute into the quotient rule: $$\frac{dy}{dx} = \frac{a\cos(ax+b)\cos(cx+d) - \sin(ax+b)(-c\sin(cx+d))}{\cos^2(cx+d)}.$$ So, $$\frac{dy}{dx} = \frac{a\cos(ax+b)\cos(cx+d) + c\sin(ax+b)\sin(cx+d)}{\cos^2(cx+d)}.$$

Answer: $$\frac{a\cos(ax+b)\cos(cx+d) + c\sin(ax+b)\sin(cx+d)}{\cos^2(cx+d)}$$