|CALCULUS I TEST #3 SOLUTIONS|
A/ Differentiate :
|1) applying the rules of logs:
|2) f (x) = ln (cos 2x + sin 2x) 5 arcsec x²|
(cos 2x) ' = 2 sin 2x, and (sin 2x) ' = 2 cos 2x
first fraction is du/u (derivative of ln u)
2nd fraction is derivative of arcsec u.
|3) g (t) = ln² (3t + 1)
using chain rule,we get:
|4) h (x) = csc (ln x) + arctan 5x arccos (x³ + 7)|
1st fraction: (csc u) ' = du csc u cot u with u = ln x.
2nd fraction: (arctan u) ' = du / (u² + 1) with u = 5x.
3rd fraction: [ arccos u] ' = du / (1 u²)½ with u = ( x³ + 7).
|5) we have y = e x ² 3 + (7) x 1/x log 3 (x³ 5x)|
1st term: exp u, base e so (e u ) ' = du e u, with u = x² 3
2nd term:exp u, base a = 7 so (a u ) ' = du a u ln a, with u = x 1/x.
3rd term: log u, base a = 3 so (log a u) ' = du /u ln a,with u = (x³ 5x)
B/ Use implicit or logarithmic differentiation to find y'.
1) y ² e 2x + xy³ = 1
Both terms are products, so 2y y' e 2x + 2e 2x y 2 + y 3 + 3xy 2 y' = 0 (divide through by y)
Solving for y' we get
2) y = (x + 4) x ² 1 .
taking ln of both sides, we get, ln y = (x 2 1) ln (x + 4)
We'll use the product rule with u = (x 2 1) and v = ln (x + 4).
Once we have y'/y, we'll multiply both sides by y to find y ' as required.
C/ Find an equation for the tangent
1) The slope of the tangent = ½, so y' = e x = ½.
This means that ½ = 1/e x, therefore e x = 2, so x = ln 2 and y = ½
The equation of the tangent is:
2) When x = 2, y = ln 2 and y' = 1/x.
The equation of the tangent is:
1) Rewrite as an algebraic expression in x: (diagram)
a) arctan (x + 1)/5 is an angle A with opposite = (x + 1), adjacent = 5.
b) arccos (7/x) is an angle B with adjacent = 7, hypotenuse = x.
2) We need equations for the tangent and normal lines to
y = arcsin (x 1) at the point P .
Since the derivative is the slope of the tangent, we find the
value of y' at x = 3/2. The derivative of arcsin u = du / (u² 1)½.
Now we write for the tangent, and for the normal.
The equation of the tangent line is
The equation of the normal line is
3) We have to analyze and graph y = e x ² , which is e u with u = x².
Step 1: Intercepts and Asymptotes
intercept: set x = 0, we get e 0 = 1; intercept is (0 , 1)
asymptote: since e x ² is positive (> 0) for all values of x, asymptote is y = 0 (x-axis)
Step 2: 1st derivative for Critical Points, Increasing and Decreasing intervals.
The derivative is du e u which is 2x (e x ² ), or in fraction form:
Step 3: 2nd derivative for Concavity and Points of Inflection.
Now we differentiate 2x (e x ² ) using the product rule and some factoring to get:
|Max: (0, 1) Min: none||Intercept: (0, 1)||Inflection Points:|
|Asymptotes: y = 0||Concave Up:
|Increasing: x < 0|
Decreasing: x > 0
Symmetry: to y-axis.
4) We know that
so, with the antidifferentiation power rule, on ( 3x 4 ), we get;
f ' (x) = x 3 + C, and since f ' (1) = 1, C = 2.
so f ' (x) = x 3 2.
Now we antidifferentiate this to get;
f (x) = ½ x 2 2x + D and substituting f (1) = ½, we find D = 3.
So f (x) = ½ x 2 2x + 3.
Calculus I MathRoom Index
(all content of the MathRoom Lessons © Tammy the Tutor; 2002 - ?).
M2 I2 ACIDS stands for:
Maximum, Minimum, Intercepts, Inflection points, Asymptotes,
Concavity, Increasing intervals, Decreasing intervals and Symmetry.
For notes on this topic see lesson on curve sketching.