MA140-Engineering Calculus

Lecture 10

October 9, 2019

Exercises

Prove that \frac{d}{dx}(cos x) = − sin x \frac{d}{dx}(e^{x}) = e^{x}

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Basic Rules of Differentiation

(1) Derivative of a Constant Function: If f has the constant value

f(x) = c, then:

\frac{df}{dx} = \frac{d}{dx}(c) = 0

(2) Power Rule for Positive Integers: If n is a positive integer, then

\frac{d}{dx}(x^{n}) = nx^{n−1}

(3) Constant Multiple Rule: If u is a differentiable function of x, and c is

a constant, then

\frac{d}{dx}(cu) = c\frac{du}{dx}

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(4) Derivative Sum Rule: If u and v are differentiable functions of x, then their sum u + v is differentiable at every point where u and v are both differentiable. At such points,

\frac{d}{dx}(u + v) = \frac{du}{dx} + \frac{dv}{dx}

(5) Derivative Product Rule: If u and v are differentiable at x, then so is their product uv, and

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

(6) Derivative Quotient Rule: If u and v are differentiable at x and if v(x) ≠ 0, then the quotient u/v is differentiable at x, and

\frac{d}{dx}(\frac{u}{v}) = \frac{v \frac{du}{dx} − u\frac{dv}{dx}}{v^{2}}

Example 1.1

Suppose that f(x) = −5x^{3} + 3x^{2} − 9x + 7, then find:

(a) The derivative of f(x)

(b) The slope of the tangent line at x = 2

(c) The equation of the tangent at x = 2

Note: The equation of a line with slope m and a point (x_{1}, y_{1}) on the line is:

y − y_{1} = m(x − x_{1})

(a):

f′(x) = −15x^{2} + 6x − 9

(b): The slope of the tangent line at x = 2 is f′(2)

f′(2) = −15(2)^{2} + 6(2) − 9 = −15(4) + 12 − 9 = −60 + 12 − 9 = −57 (c): The y coordinate at x = 2 is

f(2) = −5(2)^{3} + 3(2)^{2} − 9(2) + 7 = −5(8) + 3(4) − 18 + 7

= −40 + 12 − 18 + 7 = −39

So (2, −39) is a point on the tangent line and the slope of the line is −57 so the equation of the line is:

y − y_{1} = m(x − x_{1}) = y + 39 = −57(x − 2)

Example 1.2

Use the rules to show that

\frac{d}{dx}(tan x) = sec^{2} x

\frac{d}{dx}(tan x) = \frac{d}{dx}(\frac{sin x}{cos x})

Using the quotient rule, we will get

\frac{d}{dx}(\frac{sin x}{cos x}) = \frac{(cos x)[\frac{d}{dx}(sin x)] − (sin x)[\frac{d}{dx}(cos x)]}{(cos^{2} x)}

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Exercises

Use the rules to show that: \frac{d}{dx}(sec x) = sec(x) · tan(x) \frac{d}{dx}(csc(x)) = − csc(x) · cot(x)

Hint: csc(x) = 1/ sin(x) and cot(x) = 1/ tan(x)

\frac{d}{dx}(cot(x)) = − csc^{2}(x)

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Definition 1.3

If f(u) is differentiable at the point u = g(x) and g(x) is differentiable at x, then the composite function (f ∘ g)(x) = f(g(x)) is differentiable at x, and

(f ∘ g)′(x) = f′(g(x)) · g′(x)

In another notation, if y = f(u) and u = g(x), then

\frac{dy}{dx} = \frac{dy}{du} · \frac{du}{dx}

Example 1.4

If y = (x^{3} + 4x^{4} + 7)^{99}, find \frac{dy}{dx} Let u = x^{3} + 4x^{4} + 7, we can write y as y = u^{99}, then by chain rule we have:

\frac{dy}{dx} = \frac{dy}{du} · \frac{du}{dx} = 99u^{98}[3x^{2} + 16x^{3}],

so

\frac{dy}{dx} = 99(x^{3} + 4x^{4} + 7)^{98}(3x^{2} + 16x^{3})

Example 1.5

If y = \frac{1000}{(x^{4} + 2x^{2} + 8)^{40}} , find \frac{dy}{dx}

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y = 1000(x^{4} + 2x^{2} + 8)^{−40}. Let u = x^{4} + 2x^{2} + 8, so we can write y = 1000u^{−40} The Chain rule is:

\frac{dy}{dx} = \frac{dy}{du} · \frac{du}{dx}

Now:

\frac{dy}{du} = (1000)(−40)u^{−41}

\frac{du}{dx} = 4x^{3} + 4x

so

\frac{dy}{dx} = −40000u^{−41}[4x^{3} + 4x]

then

\frac{dy}{dx} = \frac{−40000}{u^{41}}(4x^{3} + 4x)

or

\frac{dy}{dx} = \frac{−40000(4x^{3} + 4x)}{x^{4} + 2x^{2} + 8}

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Note: Sometimes it is useful to involve a second (or more) intermediate function

\frac{dy}{dx} = \frac{dy}{du} · \frac{du}{dv} · \frac{dv}{dx}

Example 1.6

Find \frac{dy}{dx}, when

y = sin^{4}(x^{5} + 7)

Let u = sin(x^{5} + 7) and let v = x^{5} + 7 so the chain rule gives

\frac{d sin^{4}(x^{5} + 7)}{dx} = \frac{d sin^{4}(x^{5} + 7)}{d sin(x^{5} + 7)} · \frac{d sin(x^{5} + 7)}{d(x^{5} + 7)} · \frac{d(x^{5} + 7)}{dx}

[4 sin^{3}(x^{5} + 7)] · [ cos(x^{5} + 7)] · [5x^{4}] = 20x^{4} · sin^{3}(x^{5} + 7) · cos(x^{5} + 7)

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Exercises

Find \frac{dy}{dx}, if:

y = x^{2}e^{sin x}
y = tan^{3}[sin^{2}(x^{4})]
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