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Anti-derivative of (tan x – 1)/(tan x + 1) with respect to x is
To determine the antiderivative (indefinite integral) of I = ∫ (tan x - 1) / (tan x + 1) dx Step 1: Substituting x in terms of a trigonometric identity We apply the identity : tan(A - B) = (tan A - tan B) / (1 + tan A tan B) Here, we choose A = π/4 and B = x, so tan(π/4 - x) = (tan(π/4) - tan x) / (Read more
To determine the antiderivative (indefinite integral) of
I = ∫ (tan x – 1) / (tan x + 1) dx
Step 1: Substituting x in terms of a trigonometric identity
We apply the identity :
tan(A – B) = (tan A – tan B) / (1 + tan A tan B)
Here, we choose A = π/4 and B = x, so
tan(π/4 – x) = (tan(π/4) – tan x) / (1 + tan(π/4) tan x)
Since tan(π/4) = 1, this reduces to:
tan(π/4 – x) = (1 – tan x) / (1 + tan x)
Taking the negative,
– tan(π/4 – x) = (tan x – 1) / (tan x + 1)
So, our integral is:
I = ∫ – tan(π/4 – x) dx
Step 2: Finding the Integral
We know:
∫ tan u du = log | sec u | + C
Substituting u = π/4 – x, we get:
I = – ∫ tan(π/4 – x) dx
= – log | sec(π/4 – x) | + C
Conclusion
Therefore, the right answer is: – log | sec(π/4 – x) | + C
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In a sphere the rate of change of volume is
The correct answer is: surface area times the rate of change of radius. This follows because the volume V of a sphere is connected to its radius r by the formula: V = (4/3) π r³ The rate of change of volume with respect to time is: dV/dt = 4 π r² (dr/dt) Here, 4 π r² is the surface area of the spherRead more
The correct answer is: surface area times the rate of change of radius.
This follows because the volume V of a sphere is connected to its radius r by the formula:
V = (4/3) π r³
The rate of change of volume with respect to time is:
dV/dt = 4 π r² (dr/dt)
Here, 4 π r² is the surface area of the sphere, and (dr/dt) is the rate of change of the radius. Therefore, the rate of change of volume is equal to the surface area times the rate of change of radius.
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If V = 4/3 πr³, at what rate in cubic units is V increasing when r = 10 and dr/dt = 0.01?
To find the rate at which the volume V is increasing, we need to differentiate the volume formula V = (4/3) π r³ with respect to time t. dV/dt = 4 π r² (dr/dt) Now, plug in the values: r = 10 and (dr/dt) = 0.01: dV/dt = 4 π (10)² (0.01) dV/dt = 4 π × 100 × 0.01 = 4 π Thus, the rate at which the voluRead more
To find the rate at which the volume V is increasing, we need to differentiate the volume formula V = (4/3) π r³ with respect to time t.
dV/dt = 4 π r² (dr/dt)
Now, plug in the values: r = 10 and (dr/dt) = 0.01:
dV/dt = 4 π (10)² (0.01)
dV/dt = 4 π × 100 × 0.01 = 4 π
Thus, the rate at which the volume is increasing is 4π cubic units.
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The function f(x) = tan x – x
The function f(x) = tan x - x increases or decreases throughout, and the only way to find this is by analyzing the derivative of f(x). Now, differentiate f(x) with respect to x: f'(x) = d/dx(tan x) - d/dx(x) = sec² x - 1 So, f'(x) = sec² x - 1 = tan² x Since tan² x is always non-negative for all reaRead more
The function f(x) = tan x – x increases or decreases throughout, and the only way to find this is by analyzing the derivative of f(x).
Now, differentiate f(x) with respect to x:
f'(x) = d/dx(tan x) – d/dx(x) = sec² x – 1
So,
f'(x) = sec² x – 1 = tan² x
Since tan² x is always non-negative for all real values of x, f'(x) ≥ 0 for all x where tan x is defined.
Therefore, f(x) is always increasing where it is defined, but has vertical asymptotes at x = (π/2) + nπ, where n is any integer.
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For all real values of x, the function f(x) = |x| is
The function f(x) = |x| is neither increasing nor decreasing throughout its domain since: - For x ≥ 0, the function is increasing (since f(x) = x for nonnegative x). - For x < 0, the function is decreasing (since f(x) = -x for negative x). The function is neither strictly increasing nor strictlyRead more
The function f(x) = |x| is neither increasing nor decreasing throughout its domain since:
– For x ≥ 0, the function is increasing (since f(x) = x for nonnegative x).
– For x < 0, the function is decreasing (since f(x) = -x for negative x).
The function is neither strictly increasing nor strictly decreasing at x = 0 because of a sharp "corner."
Hence f(x) = |x| is neither increasing nor decreasing on its entire domain.
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