Download 22 Derivative of inverse function and more Slides Calculus in PDF only on Docsity!
22.1 Statement
Any time we have a function f , it makes sense to form is inverse function f −^1 (although this often requires a reduction in the domain of f in order to make it injective). If we know the derivative of f , then we can find the derivative of f −^1 as follows:
Derivative of inverse function. If f is a function with in- verse function f −^1 , then ( f −^1
(x) =
f ′(f −^1 (x))
This formula is an immediate consequence of the definition of an inverse function and the chain rule:
f (f −^1 (x)) = x d dx
[
f (f −^1 (x))
]
d dx
[x]
f ′(f −^1 (x)) · (f −^1 )′(x) = 1
(f −^1 )′(x) =
f ′(f −^1 (x))
The figure below shows that the formula agrees with the fact that the graph of f −^1 is the reflection across the 45◦^ line y = x of the graph of f. Such a reflection interchanges the coordinates of a point (i.e., (x, y) reflects to (y, x)), so the reflection of a line has slope the reciprocal of the slope of the original line. Thus, the slope of the line tangent to the graph of f −^1 at the point (x, f −^1 (x)) (red line) is the reciprocal of the slope of the tangent to the graph of f at the point (f −^1 (x), x) (green line), and this is also what the formula says.
22.1.1 Example The inverse of the function f (x) = x^2 with reduced do- main [0, ∞) is f −^1 (x) =
x. Use the formula given above to find the derivative of f −^1.
Solution We have f ′(x) = 2x, so that f ′(f −^1 (x)) = 2
x. Using the formula above, we have
(f −^1 )′(x) =
f ′(f −^1 (x))
x
(We can check by using the power rule:
(f −^1 )′(x) = d dx
[√
x
]
d dx
[
x^1 /^2
]
= 12 x−^1 /^2 =
x
in agreement with what we just found.)
22.2 Derivative of logarithm function
The logarithm function loga x is the inverse of the exponential function ax. Therefore, we can use the formula from the previous section to obtain its deriva- tive.
Derivative of logarithm function. For any positive real number a, d dx
[loga x] =
x ln a
In particular, d dx
[ln x] =
x
The second formula follows from the first since ln e = 1. We verify the first formula. The function f (x) = ax^ has inverse function f −^1 (x) = loga x. We
Derivatives of inverse sine and inverse cosine func- tions.
(i)
d dx
[
sin−^1 x
]
1 − x^2
(ii) d dx
[
cos−^1 x
]
1 − x^2
We verify the first formula. The function f (x) = sin x with domain reduced to [−π/ 2 , π/2] has inverse function f −^1 (x) = sin−^1 x. We have f ′(x) = cos x, so that f ′(f −^1 (x)) = cos
sin−^1 (x)
. The formula for the derivative of an inverse function now gives
d dx
[
sin−^1 x
]
= (f −^1 )′(x) =
f ′(f −^1 (x))
cos
sin−^1 (x)
This last expression can be simplified by using the trigonometric identity sin^2 θ+ cos^2 θ = 1. Put θ = sin−^1 (x) and note that θ ∈ [−π/ 2 , π/2]. Then,
cos
sin−^1 (x)
= cos θ =
1 − sin^2 θ =
1 − x^2 ,
where we have used that cos θ ≥ 0 in choosing the positive square root when we solved the trigonometric identity for cos θ. Putting this final expression into the earlier equation, we get
d dx
[
sin−^1 x
]
1 − x^2
as claimed.
22.3.1 Example Find the derivative of each of the following functions:
(a) f (x) = 4ex^ sin−^1 x,
(b) f (x) = cos−^1
x^3 + x
Solution
(a) The product rule is applied first:
f ′(x) =
d dx
[
4 ex^ sin−^1 x
]
d dx
[4ex] sin−^1 x + 4ex^ d dx
[
sin−^1 x
]
= (4ex) sin−^1 x + 4ex
1 − x^2
= 4ex^ sin−^1 x + 4 ex √ 1 − x^2
(b) The chain rule is used with outside function the arc cosine function:
f ′(x) =
d dx
[
cos−^1
x^3 + x
)]
1 − (x^3 + x)^2
· (3x^2 + 1)
3 x^2 + 1 √ 1 − (x^3 + x)^2
22 – Exercises
22 – 1 Find the derivatives of each of the following functions:
(a) f (x) = 3 log 3 x − 4 ln x,
(b) f (t) = ln
1 + 3e^2 t
22 – 2 Find the derivative of the function f (x) = ln (ln (ln x)).
22 – 3 Find the derivative of the function f (x) = sin−^1
1 − x^2
