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5128_Ch03_pp098-184

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Section 3.5 Derivatives of Trigonometric Functions 145<br />

d<br />

tan x sec d x<br />

2 x,<br />

d<br />

cot x csc d x<br />

2 x,<br />

d<br />

sec x sec x tan x<br />

d x<br />

d<br />

csc x csc x cot x<br />

d x<br />

EXAMPLE 4<br />

Finding Tangent and Normal Lines<br />

Find equations for the lines that are tangent and normal to the graph of<br />

f x tan x<br />

<br />

x<br />

at x 2. Support graphically.<br />

SOLUTION<br />

Solve Numerically Since we will be using a calculator approximation for f 2 anyway,<br />

this is a good place to use NDER.<br />

We compute tan 22 on the calculator and store it as k. The slope of the tangent line<br />

at 2, k is<br />

NDER ( tan x<br />

,<br />

x<br />

2) ,<br />

y 1 = tan (x)/ x<br />

y 2 = 3.43x – 7.96<br />

y 3 = –0.291x – 0.51<br />

which we compute and store as m. The equation of the tangent line is<br />

y k mx 2, or<br />

y mx k 2m.<br />

Only after we have found m and k 2m do we round the coefficients, giving the<br />

tangent line as<br />

y 3.43x 7.96.<br />

The equation of the normal line is<br />

1<br />

y k x 2, or<br />

m<br />

y m<br />

1 x k m<br />

2 .<br />

X=2<br />

Y=–1.09252<br />

[–3/2, 3/2] by [–3, 3]<br />

Figure 3.40 Graphical support for<br />

Example 4.<br />

Again we wait until the end to round the coefficients, giving the normal line as<br />

y 0.291x 0.51.<br />

Support Graphically Figure 3.40, showing the original function and the two lines,<br />

supports our computations. Now try Exercise 23.<br />

EXAMPLE 5<br />

Find y if y sec x.<br />

SOLUTION<br />

A Trigonometric Second Derivative<br />

y sec x<br />

ysec x tan x<br />

d<br />

ysec x tan x<br />

d x<br />

d<br />

d<br />

sec x tan x tan x sec x<br />

d x<br />

d x<br />

sec x sec 2 x tan x sec x tan x<br />

sec 3 x sec x tan 2 x Now try Exercise 36.

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