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1 Lecture 18: Inverse functions, the derivative of
ln(x).
1.1 Outline
- The derivative of an inverse function
- The derivative of ln(x).
- Derivatives of inverse trigonometric functions
1.2 The graph of inverse function
We consider the graph of a function f and let (a, f (a)) = (a, b) be a point on the graph. We recall that if we graph a function f which is one-to-one, we may find the graph of the inverse function by interchanging the x and y coordinates. Thus (b, a) = (b, f −^1 (b)) will be a point on the graph of f −^1. If m is the slope of the tangent line to the graph of f at (a, b), we may find the tangent line to the graph at f −^1 at (b, a), we interchange the rise and the run and find that the slope is 1/m. Thus, we may find the derivative of the inverse function in terms of the derivative of
f as
(f −^1 )′(b) =
f ′(f −^1 (b))
This will hold provide f is one-to-one, differentiable at f −^1 (b) and f ′(f −^1 (b)) is not zero. The formula for the derivative of an inverse function (1) may seem rather complicated, but it helps to remember that the tangent line to the graph of f −^1 at b corresponds to the tangent line of the graph of f at a = f −^1 (b). We will see that the formula is easy to use to find find derivatives of the logarithm and inverse trig functions.
Example. The function f is one-to-one and differentiable. The tangent line to the graph of y = f (x) at x = 2 is y = 3x + 2. Find the tangent line to the graph of f −^1 at x = 8.
Solution. Since y = 3x+2 is the tangent line to the graph of f , it passes through the point (2, f (2)) and has slope f ′(2). Thus, we have f (2) = 3 · 2 + 2 = 8 and f ′(2) = 3. Since f (2) = 8, we know f −^1 (8) = 2 and from (1), we have that
(f −^1 )′(8) =
f ′(2)
Thus the tangent line to the graph of f −^1 at 8 is y − 2 = 13 (x − 8) which simplifies to
y =
x −
1.3 Some useful derivatives
We recall that the natural logarithm, ln(x) is defined as the inverse of the exponential function ex. If we use (1) with f (x) = ex^ and f −^1 (x) = ln(x), we obtain
d dx
ln(x) =
eln(x)^
x
Where we have used that eln(x)^ = x and (^) dxd ex^ = ex.
Example. Find the derivative of
f (x) = x ln(x) − x.
1.4 Inverse trigonometric functions
Finally, we find the derivatives of the inverse trigonometric functions. For these functions, we will need to use trigonometric identities to simplify the result of (1). We begin by finding the derivative of sin−^1 (x). This function is often written as arcsin, but we will not use this notation in this course. Please remember that sin−^1 is in the inverse function. The reciprocal or multiplicative inverse is 1/sin(x) = csc(x). If f (x) = sin(x) and f −^1 (x) = sin−^1 (x), then using our formula for the inverse function (1) we have d dx
sin−^1 (x) =
cos(sin−^1 (x))
We have a problem that we need to simplify cos(sin−^1 (x)). We recall that the range of sin−^1 is the interval [−π/ 2 , π/2] and that cos(θ) ≥ 0 for θ ∈ [−π/ 2 , π/2]. If we solve the identity sin^2 (θ) + cos^2 (θ) = 1 for cos(θ) and use that cos(θ) ≥ 0, we obtain that cos(sin−^1 (x) =
√ 1 − sin^2 (sin−^1 (x)) =
1 − x^2.
Using this we obtain that
d dx
sin−^1 (x) =
1 − x^2
, − 1 < x < 1. (2)
We cannot use (1) when x = ±1 since the denominator is zero and in fact sin−^1 is not differentiable for x = ±1. A second approach to the simplification is to draw a right triangle with angle θ = sin−^1 (x), opposite side x and hypotenuse 1. Then Pythgoras’s theorem gives us that cos(θ) =
1 − x^2.
1−x
θ
(^1) x
Example. Use (1) to find the derivative of tan−^1 (x).
Solution. We will need the identity
1 + tan^2 (x) = sec^2 (x). (3)
This may be established by dividing both sides of the identity sin^2 (x) + cos^2 (x) = 1 by cos^2 (x).
Since the derivative of tan(x) is sec(x), we have from (1) that
d dx
tan−^1 (x) =
sec^2 (tan−^1 (x))
We use the identity (3) to simplify the denominator sec^2 (tan−^1 (x)) = 1+sec^2 (tan−^1 (x)) = 1 + x^2 and find that d dx
tan−^1 (x) =
1 + x^2
Exercise. Find the derivatives of the remaining inverse trigonometric functions,
d dx
cot−^1 (x) =
1 + x^2
d dx
csc−^1 (x) =
|x|
x^2 − 1
d dx
sec−^1 (x) =
|x|
x^2 − 1
d dx
cos−^1 (x) =
1 − x^2
Of these, we will only need to remember the derivative of sec−^1 (x).
Example. Find the derivative of sin−^1 (1/x) and show that
d dx
sin−^1 (1/x) =
d dx
csc−^1 (x).
Why is this?
Solution. Since csc(θ) = 1/ sin(θ), if x = csc(θ), then 1/x = sin(θ) which implies that sin−^1 (1/x) = csc−^1 (x). Thus the derivatives are equal. To compute the derivative, we use the chain rule
d dx
sin−^1 (1/x) =
x^2
√ 1 − (1/x)^2
x^2
√ 1 /x^2
x^2 − 1
|x|
x^2 − 1
Example. Suppose that one leg of a right triangle is fixed at 10 meters and the other leg is decreasing at a rate of 0.7 meters/second. Find the rate of change of the angle opposite the fixed leg when the length of the varying leg is 20 meters.
