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[h]SL 5.6 Derivatives of xn , sinx , cosx , tanx

[q] **Differentiation Rules**

In this section, we will drastically expand our repertoire for what functions we can differentiate, and what combinations of functions we can differentiate. Again, we could derive these rules from ‘first principles’ (using limits), but that is not necessary in SL.

[q]

OTHER RULES:

\[

f(x) = x^n \Rightarrow f'(x) = nx^{n-1}

f(x) = e^x \Rightarrow f'(x) = e^x

f(x) = \ln x \Rightarrow f'(x) = \frac{1}{x}

f(x) = \sin x \Rightarrow f'(x) = \cos x

f(x) = \tan x \Rightarrow f'(x) = \sec^2 x

\]

[a]

IMPLICATIONS:

Just a reminder—this means if you wanted to know the slope of the tangent of \(y = \sin x\) at \(x = \frac{\pi}{2}\), we get \(dy = \cos x\), then do \(\cos(\frac{\pi}{2}) = 0\). So \(y = \sin x\) has a slope of 0 at that point.

[q]

NOTE: When adding terms/functions, the derivative is simply the sum of each term’s derivative. Also, if a term is multiplied by a constant, then its derivative is also just multiplied by that same constant, i.e.:

\[

\frac{d}{dx}(f(x) + g(x)) = f'(x) + g'(x) \quad \text{and} \quad \frac{d}{dx}(af(x)) = af'(x)

\]

Example : \(f(x) = 15x^4 + 3\)

\(\Rightarrow f'(x) = 60x^3\)

Example : \(y = 5\cos x\)

\(\Rightarrow \frac{dy}{dx} = -5\sin x\)

Example : \(g(x) = e^{x^2 – 4}\)

\(\Rightarrow g'(x) = 2xe^{x^2 – 4}\)

[a]

PRODUCT RULE:

So far, we have only seen rules for when we have one function by itself as well as addition of functions, and multiplication by a constant. What about the product of two functions?

\[

\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)

\]

Example 4: If \(f(x) = (x^2 + 3x)(2x – 1)\), find \(f'(x)\):

1. \(u = x^2 + 3x\)

2. \(v = 2x – 1\)

3. \(u’ = 2x + 3\)

4. \(v’ = 2\)

\[

f'(x) = (x^2 + 3x)(2) + (2x – 1)(2x + 3) = 6x^2 + 10x – 3

\]

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QUOTIENT RULE:

The next logical step would be to consider what happens when we have a function divided by another function.

Again, writing out \(u, u’, v, v’\) will help us organize things.

Example 6: Differentiate \(g(x) = \frac{5x – 1}{3x^2}\):

1. \(u = 5x – 1\)

2. \(v = 3x^2\)

3. \(u’ = 5\)

4. \(v’ = 6x\)

Using the quotient rule:

\[

g'(x) = \frac{(3x^2)(5) – (5x – 1)(6x)}{(3x^2)^2}

\]

[a]

CHAIN RULE:

The next step, after products and quotients of functions, is to deal with composite functions, i.e., how to differentiate \(f(g(x))\):

\[

y = g(u), \text{ where } u = f(x)

\]

This is a fairly confusing way of presenting the chain rule, so the first time you learn it, the following may be easier:

\[

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

\]

So when you need to differentiate functions for any reason, you must be careful not to overlook a situation where the chain rule needs to be used—students frequently overlook this.

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