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Francesco Giannino » 6.Differentiation. Derivative rules, the chain rule.


Derivative Rules

Remembering the limit laws, we recall the following theorem.

Theorem 1. Suppose that functions f and g are defined on some open interval (a, b) and f ‘(x) and g ‘(x) exist at each point x in (a, b). Then

\begin{itemize}

(f+g)'(x)=f'(x)+g'(x)\,\,\,\,\,\text{(The Sum Rule)}

\item $(f-g)'(x)=f'(x)-g'(x)\,\,\,\,\,\text{(The Difference Rule)}$

\item $(kf)'(x)=kf'(x)\,\,\,\,\,\text{for each constant $k$}\,\,\,\,\,\text{(The Multiple Rule)}$

\item $(f\cdot g)'(x)=f'(x)\cdot g(x)+f(x)\cdot g'(x) \,\,\,\,\,\text{(The Product Rule)}$

( \frac f g )^{'}=\frac{f^{'}(x) \cdot g(x)-f(x) \cdot g^{'}(x)}{g^2(x)}\,\;\text{(The Quotient Rule)}

Derivative Rules

To emphasize the fact that the derivatives are taken with respect to the independent variable x, we use the following notation, as is customary:
$$f'(x)=\frac{d}{dx}(f(x))$$

Derivative Rules

Based on previous Theorem 1. and the definition of the derivative, we get the following theorem.

Theorem 2. Suppose that functions f and g are defined on some open interval (a, b) and f ‘(x) and g ‘(x) exist at each point x in (a, b). Then
\begin{itemize}

  • \frac{d(k)}{dx}=0\;\; \text{for each real constant k}
  • \frac d {dx}(x^n)= nx^{n-1}\;\;\text {for each real number}\;\; x\;\; \text{and natural number }\;n
  • \frac d {dx}(log_ax)= \frac 1 x log_a e\;\;\text {for each real number }a>0\;\; and\;\; {a \neq 1}
  • \frac d {dx}(ln\;x)= nx^{n-1}\;\;\text {for each real number} x

Derivative Rules

\begin{itemize}

  • \frac d {dx}(a^x)\;\;\text {for each real number}\; x\;\text {and} \; a>0\; \text {and} \; a\neq 1
  • \frac d {dx}(e^x)=e^x\; \text{for each real number}\;x
  • \frac d {dx}(sin\;x)=cos\;x\;\text{for each real number}\;x
  • \frac d {dx}(cos\;x)=-sin\;x\;\text{for each real number}\;x
  • \frac d {dx}(tan\;x)=\frac 1 {cos^2x}\;x\;\text{for each real number}\;x

Exercises

Exercise 1. Compute the following derivative:

$$\frac{d}{dx}(4x^{3} - 3x^{2} + 2x + 10)$$

Solution. Using the sum, difference and constant multiple rules, we get:

$$\frac{d}{dx}(4x^{3} - 3x^{2} + 2x + 10)=$$
$$= 4\frac{d}{dx}(x^{3})- 3\frac{d}{dx}(x^{2})+ 2\frac{d}{dx}(x)+ 0=$$
$$=4\cdot 3x^{2} - 3\cdot 2x +2\cdot 1= 12x^{2} - 6x + 2.$$

Exercises

Exercise 2. Compute the following derivative:
$$\frac{d}{dx}(4\sin x - 3\cos x)$$
Solution. Using the sum, difference and constant multiple rules, we get:
$$\frac{d}{dx}(4\sin x - 3\cos x)=$$
$$=4\frac{d}{dx}(\sin x) - 3\frac{d}{dx}(\cos x)=$$
$$=4\cdot \cos x - 3\cdot (-\sin x)=4\cos x +3\sin x$$

Exercises

Exercise 3. Compute the following derivative:
$$\frac{d}{dx}(x\sin x + x^{2}\cos x).$$

Solution. Using the sum and product rules, we get:
$$\frac{d}{dx}(x\sin x + x^{2}\cos x)=$$
$$=\frac{d}{dx}(x\sin x)+\frac{d}{dx}(x^{2}\cos x)=$$
$$=\sin x + x\cos x + 2x\cos x - x^{2}\sin x=$$
$$=\sin x + 3x\cos x - x^{2}\sin x$$

Exercises

Exercise 4. Compute the following derivative:
$$\frac{d}{dx}(\frac{x^{3}+1}{x^{2}+4}).$$
Solution. Using the sum and quotient rules, we get:
$$\frac{d}{dx}(\frac{x^{3}+1}{x^{2}+4})=$$
$$=\frac{3x^{2}(x^{2}+4)-(x^{3}+1)2x}{(x^{2}+4)^{2}}=
\frac{3x^{4}+ 12x^{2}- 2x^{4} - 2x}{(x^{2}+4)^{2}}=$$
$$=\frac{x^{4}+ 12x^{2} - 2x}{(x^{2}+4)^{2}}.$$

Exercises

Exercise 5. Suppose $$f(x)=x^{3}$$ Use the definition of derivative to find f’(x).
Solution. Using the definition of derivative, we get:
$$\frac{d}{dx}(x^{3})= \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}=$$
$$=\lim_{h\rightarrow 0}\frac{(x+h)^{3}-x^{3}}{h}=\lim_{h\rightarrow 0}\frac{x^{3} + h^{3}+3hx^{2}+3xh^{2}-x^{3}}{h}=$$
$$=\lim_{h\rightarrow 0}\frac{h^{3}+3hx^{2}+3xh^{2}}{h}==\lim_{h\rightarrow 0}\frac{h\cdot (h^{2}+3x^{2}+3xh)}{h}=$$
$$=\lim_{h\rightarrow 0}(h^{2}+3x^{2}+3xh)= 3x^{2}$$
$$\Rightarrow f'(x)=3x^{2}.$$

Exercises

Exercise 6. Suppose $$f(x)=\sqrt{x}.$$ Use the definition of derivative to find f’(x).
Solution. Using the definition of derivative, we get:
$$\frac{d}{dx}(\sqrt{x})=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}=$$
$$=\lim_{h\rightarrow 0}\frac{\sqrt{x+h}- \sqrt{x}}{h}=\lim_{h\rightarrow 0}\frac{\sqrt{x+h}- \sqrt{x}}{h}\cdot \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}}=$$
$$=\lim_{h\rightarrow 0}\frac{x+h-x}{h \cdot (\sqrt{x+h} + \sqrt{x})}=
\lim_{h\rightarrow 0}\frac{h}{h \cdot (\sqrt{x+h} + \sqrt{x})}=$$$$=\lim_{h\rightarrow 0}\frac{1}{(\sqrt{x+h} + \sqrt{x})}=\frac{1}{2\sqrt{x}}.$$
$$\Rightarrow f'(x)=\frac{1}{2\sqrt{x}}.$$

The Chain Rule

Suppose we have two functions, u and y, related by the equations: u = g (x) and y = f (u). Then y = (f ° g)(x) = f (g (x)).
The chain rule deals with the derivative of the composition and may be stated as the following theorem:
Theorem 3. (The Chain Rule). Suppose that g is defined in an open interval I containing c, and f is defined in an open interval J containing g(c), such that g(x) is in J for all x in I. If g is differentiable at c, and f is differentiable at g(c), then the composition (f ° g) is differentiable at c and
$$f(g(c))'=f'(g(c))\cdot g'(c).$$

Exercises

Exercise 7. Let $f(x)=x^{2}+1 \,\,\, \text{and}\,\,\, g(x)=x^{3}+4.$ Find (f _ g)’(x).
Solution. Using the chain rule to derive, we get:
$$f(g(x))=(x^{3}+4)^{2}+1 \,\,\, \text{and}\,\,\,f(g(x))'=f'(g(x))\cdot g'(x)$$
$$\Rightarrow f(g(x))'=2(x^{3}+4)^{1}\cdot 3x^{2}=$$
$$=6x^{2}(x^{3}+4).$$

Exercises

Exercise 8. Let $$f(x)=\sin(x^{2}+3).$$ Find f’(x).
Solution. Using the chain rule, we get:
$$f'(x)=\cos(x^{2}+3)\cdot (2x)=$$
$$=2x\cos(x^{2}+3).$$

Exercises

Exercise 9. Let $$f(x)=(\sin(4x+1)+3)^{2}.$$ Find f’(x).
Solution. Using the chain rule, we get:
$$f'(x)=2(\sin(4x+1)+3)^{1}\cdot \cos(4x+1) \cdot 4=$$
$$=8(\sin(4x+1)+3)\cos(4x+1).$$

Exercises

Exercise 10. Let $$f(x)=\cos(3x+1)^{5}.$$ Find f’(x).
Solution. Using the chain rule, we get:
$$f'(x)=5 \cdot \cos(3x+1)^{4} \cdot (-\sin(3x+1)) \cdot 3=$$
$$=-15 \cos(3x+1)^{4}\sin(3x+1).$$

Equation of the line tangent to the graph of f

Recall that the line tangent to the graph of a function f at (c, f(c)) has slope f ‘(c) and has equation:
$$y-f(c)=f'(c)(x-c)$$

Exercises

Exercise 11. Find the equation of the tangent line for the graph of the following function f at the given point c:
$$f(x)=x^{3}+4x-12,\,\,c=1$$
Solution. We calculate the function f at the given point c=1:
$$f(c)=f(1)=1+4-12=-7$$

Then using the derivative rules, we get:
$$f'(x)=3x^{2}+4 \Rightarrow[tex]<br />
[tex]f'(c)=f'(1)=3\cdot 1+4$$
$$\Rightarrow f'(1)=7$$
$$\text{The equation of tangent line is:}\,\,\,y+7=7(x-1)$$

Exercises

Exercise 12. Find the equation of the tangent line for the graph of the following function f at the given point c:
$$f(x)=3\sin x +4\cos x,\,\,c=0$$
Solution. We calculate the function f at the given point c=0:
$$f(c)=f(0)=3\cdot 0 +4\cdot 1 = 4$$
Then using the derivative rules, we get:
$$f'(x)=3\cos x -4\sin x \Rightarrowf'(c)=f'(0)=3\cdot 1 -4 \cdot 0$$
$$\Rightarrow f'(0)=3$$
$$\text{The equation of tangent line is:}\,\,\,y-4=3x$$

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