Rules of Differentiation

Let \(u\) and \(v\) be differentiable at \(x\). In the numerator of the rate of change, we subtract and add the same term \(u(x)v(x+h)\): $$\begin{aligned} \dfrac{f(x+h)-f(x)}{h} &= \dfrac{u(x+h)v(x+h) - u(x)v(x)}{h} \\ &= \dfrac{u(x+h)v(x+h) - u(x)v(x+h) + u(x)v(x+h) - u(x)v(x)}{h} \\ &= \dfrac{u(x+h)-u(x)}{h}\times v(x+h) + u(x)\times \dfrac{v(x+h)-v(x)}{h}. \end{aligned}$$ Now let \(h\) approach \(0\):

• \(\dfrac{u(x+h)-u(x)}{h} \xrightarrow[h \to 0]{} u'(x)\) and \(\dfrac{v(x+h)-v(x)}{h} \xrightarrow[h \to 0]{} v'(x)\) (definition of the derivative);
• \(v(x+h) \xrightarrow[h \to 0]{} v(x)\), because \(v(x+h) - v(x) = h \times \dfrac{v(x+h)-v(x)}{h} \xrightarrow[h \to 0]{} 0 \times v'(x) = 0\).

By the rules on limits (product and sum), the rate of change tends to \(u'(x)v(x) + u(x)v'(x)\), which proves \(f'(x) = u'(x)v(x) + u(x)v'(x)\).

Let \(x \in I\) and \(h \neq 0\) small enough that \(x + h \in I\); then \(v(x+h) \neq 0\) and $$\begin{aligned} \dfrac{f(x+h)-f(x)}{h} &= \dfrac{\frac{1}{v(x+h)} - \frac{1}{v(x)}}{h} \\ &= \dfrac{v(x) - v(x+h)}{h\, v(x)\, v(x+h)} && \text{(common denominator)} \\ &= -\dfrac{v(x+h)-v(x)}{h} \times \dfrac{1}{v(x)\,v(x+h)}. \end{aligned}$$ As \(h\) approaches \(0\), the first factor tends to \(v'(x)\) (definition of the derivative) and \(v(x+h)\) tends to \(v(x)\), because \(v(x+h) - v(x) = h \times \dfrac{v(x+h)-v(x)}{h} \xrightarrow[h \to 0]{} 0 \times v'(x) = 0\). Hence the rate of change tends to \(-v'(x) \times \dfrac{1}{v(x)\,v(x)} = -\dfrac{v'(x)}{[v(x)]^2}\).

We write the quotient as a product with a reciprocal: \(f(x) = \dfrac{u(x)}{v(x)} = u(x) \times \dfrac{1}{v(x)}\). By the reciprocal rule, \(\left(\dfrac{1}{v}\right)'(x) = -\dfrac{v'(x)}{[v(x)]^2}\). Applying the product rule: $$\begin{aligned} f'(x) &= u'(x) \times \dfrac{1}{v(x)} + u(x) \times \left(-\dfrac{v'(x)}{[v(x)]^2}\right) \\ &= \dfrac{u'(x)}{v(x)} - \dfrac{u(x)v'(x)}{[v(x)]^2} \\ &= \dfrac{u'(x)v(x)}{[v(x)]^2} - \dfrac{u(x)v'(x)}{[v(x)]^2} && \text{(common denominator)} \\ &= \dfrac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}. \end{aligned}$$

If \(a = 0\), then \(f(x) = g(b)\) is constant, so \(f'(x) = 0 = a \times g'(b)\).
Assume \(a \neq 0\) and set \(X = ax+b\). For \(h \neq 0\), \(f(x+h) = g(a(x+h)+b) = g(X + ah)\). Let \(k = ah\): then \(k \neq 0\), and \(k\) tends to \(0\) when \(h\) tends to \(0\). Hence $$ \dfrac{f(x+h)-f(x)}{h} = \dfrac{g(X+k)-g(X)}{h} = \dfrac{g(X+k)-g(X)}{k} \times \dfrac{k}{h} = a \times \dfrac{g(X+k)-g(X)}{k} \xrightarrow[h \to 0]{} a \times g'(X) = a \times g'(ax+b). $$

Using the limit definition of the derivative:$$ \dfrac{\mathrm{d}}{\mathrm{d}x}(e^x) = \lim_{h \to 0} \dfrac{e^{x+h} - e^x}{h} = \lim_{h \to 0} \dfrac{e^x e^h - e^x}{h} = \lim_{h \to 0} e^x \left(\dfrac{e^h - 1}{h}\right) $$ Since \(e^x\) does not depend on \(h\), we can move it outside the limit:$$ = e^x \cdot \lim_{h \to 0} \left(\dfrac{e^h - 1}{h}\right) $$ The limit \(\lim_{h \to 0} \frac{e^h-1}{h}\) is a fundamental limit which is equal to 1.
Therefore, \(\dfrac{\mathrm{d}}{\mathrm{d}x}(e^x) = e^x \cdot 1 = e^x\).

Let the outer function be \(v(u) = e^u\) and the inner function be \(u(x)\).
Then, \(e^{u(x)} = v(u(x))\).

• The derivative of the outer function \(v(u) = e^u\) with respect to \(u\) is \(v'(u) = e^u\).
• The derivative of the inner function \(u(x)\) with respect to \(x\) is \(u'(x)\).

As chain rule formula:$$\begin{aligned} \left(e^{u(x)}\right)' &= v'(u(x)) \cdot u'(x)\\ &=e^{u(x)} \cdot u'(x) \\ \end{aligned}$$

$$\begin{aligned}\dfrac{\mathrm{d}}{\mathrm{d}x}(a^x) &= \dfrac{\mathrm{d}}{\mathrm{d}x}(e^{x\ln(a)})&(a^x = e^{\ln\left(a^x\right)} = e^{x\ln(a)})\\ &= e^{x\ln(a)}\dfrac{\mathrm{d}}{\mathrm{d}x}\left(x\ln(a)\right) \\ &= a^x\cdot \ln(a) \\ \end{aligned}$$

Let \(y = \ln x\). By the definition of the natural logarithm, this means \(e^y = x\). We can now differentiate this implicit equation with respect to \(x\), using the chain rule on the left side:$$\begin{aligned}\dfrac{\mathrm{d}}{\mathrm{d}x}(e^y) &= \dfrac{\mathrm{d}}{\mathrm{d}x}(x) \\ e^y \cdot \dfrac{\mathrm{d}y}{\mathrm{d}x} &= 1 \\ \dfrac{\mathrm{d}y}{\mathrm{d}x} &= \dfrac{1}{e^y}\end{aligned}$$ Since we know that \(e^y = x\), we substitute this back into the expression:$$ \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{x} $$

We use the change of base formula to express the function in terms of the natural logarithm, and then differentiate.$$\begin{aligned}\dfrac{\mathrm{d}}{\mathrm{d}x}(\log_a x) &= \dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{\ln x}{\ln a}\right) & (\text{change of base formula}) \\ &= \dfrac{1}{\ln a} \cdot \dfrac{\mathrm{d}}{\mathrm{d}x}(\ln x) & (\text{since } \ln a \text{ is a constant}) \\ &= \dfrac{1}{\ln a} \cdot \dfrac{1}{x} \\ &= \dfrac{1}{x\ln a}\end{aligned}$$

For \(f(x) = \sin(x)\), the limit definition of the derivative, the addition formula \(\sin(x+h) = \sin x \cos h + \cos x \sin h\) and the two limits above give$$\begin{aligned}\dfrac{\sin(x+h)-\sin(x)}{h} &= \dfrac{\sin(x)\cos(h) + \cos(x)\sin(h) - \sin(x)}{h} &&\text{(addition formula)} \\ &= \sin(x)\cdot\dfrac{\cos(h)-1}{h} + \cos(x)\cdot\dfrac{\sin(h)}{h} \\ &\xrightarrow[h \to 0]{} \sin(x)\cdot 0 + \cos(x)\cdot 1 = \cos(x). &&\text{(fundamental limits)}\end{aligned}$$ For \(f(x) = \cos(x)\), the addition formula \(\cos(x+h) = \cos x \cos h - \sin x \sin h\) and the two limits above give$$\begin{aligned}\dfrac{\cos(x+h)-\cos(x)}{h} &= \cos(x)\cdot\dfrac{\cos(h)-1}{h} - \sin(x)\cdot\dfrac{\sin(h)}{h} &&\text{(addition formula)} \\ &\xrightarrow[h \to 0]{} \cos(x)\cdot 0 - \sin(x)\cdot 1 = -\sin(x). &&\text{(fundamental limits)}\end{aligned}$$

We use the quotient rule on \(f(x) = \tan x = \dfrac{\sin x}{\cos x}\).
Let \(u=\sin x\) and \(v=\cos x\). Then \(u'=\cos x\) and \(v'=-\sin x\).$$\begin{aligned}f'(x) &= \dfrac{u'v - uv'}{v^2}\\ &= \dfrac{(\cos x)(\cos x) - (\sin x)(-\sin x)}{(\cos x)^2} \\ &= \dfrac{\cos^2 x + \sin^2 x}{\cos^2 x}\\ &= \dfrac{\cos^2 x}{\cos^2 x} + \dfrac{\sin^2 x}{\cos^2 x} \\ &= 1+ \dfrac{\sin^2 x}{\cos^2 x}\\ &= 1+\tan^2 x \\ \end{aligned}$$

Proof for \(f(x) = \sec x\):We write \(f(x) = \dfrac{1}{\cos x}\) and use the quotient rule with \(u=1\) and \(v=\cos x\). Then \(u'=0\) and \(v'=-\sin x\).$$\begin{aligned}f'(x) &= \dfrac{u'v - uv'}{v^2} = \dfrac{(0)(\cos x) - (1)(-\sin x)}{(\cos x)^2} \\ &= \dfrac{\sin x}{\cos^2 x} = \dfrac{1}{\cos x} \cdot \dfrac{\sin x}{\cos x} \\ &= \sec x \tan x\end{aligned}$$ The proofs for \(\csc x\) and \(\cot x\) are similar.

Proof for \(y = \arcsin x\):If \(y = \arcsin x\), then \(\sin y = x\). We differentiate implicitly with respect to \(x\):$$\begin{aligned}\dfrac{\mathrm{d}}{\mathrm{d}x}(\sin y) &= \dfrac{\mathrm{d}}{\mathrm{d}x}(x) \\ \cos y \cdot \dfrac{\mathrm{d}y}{\mathrm{d}x} &= 1 \\ \dfrac{\mathrm{d}y}{\mathrm{d}x} &= \dfrac{1}{\cos y}\end{aligned}$$ Since \(y = \arcsin x\) lies in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), \(\cos y \geq 0\), so the identity \(\cos^2 y + \sin^2 y = 1\) gives \(\cos y = \sqrt{1-\sin^2 y}\). Since \(\sin y = x\), we have \(\cos y = \sqrt{1-x^2}\), which is not zero for \(-1 < x < 1\): the division by \(\cos y\) above is then allowed.
Substituting this back gives \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{\sqrt{1-x^2}}\).