The graph of an inverse function is the reflection of the original in the line
y = x y = x y = x f ( x ) = x 2 f(x) = x^2 f ( x ) = x 2 x ≥ 0 x \ge 0 x ≥ 0 g ( x ) = x g(x) = \sqrt{x} g ( x ) = x ( 2 , 4 ) (2, 4) ( 2 , 4 ) f f f ( 4 , 2 ) (4, 2) ( 4 , 2 ) g g g
Reflected graphs have reciprocal slopes
The curve y = x squared for x from 0 to 3 and its inverse y = square root of x, mirror images across the dashed line y = x. The tangent to y = x squared at (2, 4) has slope 4; the tangent to y = square root of x at (4, 2) has slope 1/4.
1 2 3 4 5 6 1 2 3 4 5 6 x y
(2, 4)
(4, 2)
Reflected graphs have reciprocal slopes
At ( 2 , 4 ) (2, 4) ( 2 , 4 ) x 2 x^2 x 2 4 4 4 ( 4 , 2 ) (4, 2) ( 4 , 2 ) x \sqrt{x} x 1 2 4 = 1 4 \tfrac{1}{2\sqrt{4}} = \tfrac{1}{4} 2 4 1 = 4 1
If g g g f f f f ( g ( x ) ) = x f(g(x)) = x f ( g ( x )) = x
f ′ ( g ( x ) ) ⋅ g ′ ( x ) = 1 ⟹ g ′ ( x ) = 1 f ′ ( g ( x ) ) f'\big(g(x)\big) \cdot g'(x) = 1 \quad\Longrightarrow\quad g'(x) = \frac{1}{f'\big(g(x)\big)} f ′ ( g ( x ) ) ⋅ g ′ ( x ) = 1 ⟹ g ′ ( x ) = f ′ ( g ( x ) ) 1 The slope of the inverse at x x x f f f matching point, g ( x ) g(x) g ( x )
Reflecting in y = x y = x y = x 4 4 4 1 1 1 4 4 4 1 1 1 1 4 \tfrac{1}{4} 4 1
The chain rule says the same thing about rates. Going through f f f g g g 1 1 1 Undoing a function undoes its rate of change,
so the inverse’s slope is one over the original’s. (Where f ′ f' f ′ 0 0 0
Example 1 — Without a formula for the inverse
Let f ( x ) = x 3 + x f(x) = x^3 + x f ( x ) = x 3 + x g = f − 1 g = f^{-1} g = f − 1 g ′ ( 2 ) g'(2) g ′ ( 2 )
There is no convenient formula for g g g f ( 1 ) = 2 f(1) = 2 f ( 1 ) = 2 g ( 2 ) = 1 g(2) = 1 g ( 2 ) = 1 f ′ ( x ) = 3 x 2 + 1 f'(x) = 3x^2 + 1 f ′ ( x ) = 3 x 2 + 1 f ′ ( 1 ) = 4 f'(1) = 4 f ′ ( 1 ) = 4
g ′ ( 2 ) = 1 f ′ ( g ( 2 ) ) = 1 f ′ ( 1 ) = 1 4 g'(2) = \frac{1}{f'\big(g(2)\big)} = \frac{1}{f'(1)} = \frac{1}{4} g ′ ( 2 ) = f ′ ( g ( 2 ) ) 1 = f ′ ( 1 ) 1 = 4 1 y = ln x y = \ln x y = ln x e y = x e^y = x e y = x
e y d y d x = 1 ⟹ d y d x = 1 e y = 1 x e^y \frac{dy}{dx} = 1 \quad\Longrightarrow\quad \frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x} e y d x d y = 1 ⟹ d x d y = e y 1 = x 1 For any base b > 0 b > 0 b > 0 b ≠ 1 b \ne 1 b = 1 log b x = ln x ln b \log_b x = \tfrac{\ln x}{\ln b} log b x = l n b l n x
d d x ln x = 1 x , d d x log b x = 1 x ln b \frac{d}{dx}\ln x = \frac{1}{x}, \qquad \frac{d}{dx}\log_b x = \frac{1}{x \ln b} d x d ln x = x 1 , d x d log b x = x ln b 1 y = arcsin x y = \arcsin x y = arcsin x sin y = x \sin y = x sin y = x − π 2 ≤ y ≤ π 2 -\tfrac{\pi}{2} \le y \le \tfrac{\pi}{2} − 2 π ≤ y ≤ 2 π cos y d y d x = 1 \cos y \, \tfrac{dy}{dx} = 1 cos y d x d y = 1 cos y ≥ 0 \cos y \ge 0 cos y ≥ 0 cos y = 1 − sin 2 y = 1 − x 2 \cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2} cos y = 1 − sin 2 y = 1 − x 2
d d x arcsin x = 1 1 − x 2 , d d x arccos x = − 1 1 − x 2 , d d x arctan x = 1 1 + x 2 \frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}}, \qquad
\frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1 - x^2}}, \qquad
\frac{d}{dx}\arctan x = \frac{1}{1 + x^2} d x d arcsin x = 1 − x 2 1 , d x d arccos x = − 1 − x 2 1 , d x d arctan x = 1 + x 2 1 For arctan, tan y = x \tan y = x tan y = x sec 2 y d y d x = 1 \sec^2 y \, \tfrac{dy}{dx} = 1 sec 2 y d x d y = 1 sec 2 y = 1 + tan 2 y = 1 + x 2 \sec^2 y = 1 + \tan^2 y = 1 + x^2 sec 2 y = 1 + tan 2 y = 1 + x 2
Example 2 — With the chain rule
Differentiate y = arctan ( 3 x ) y = \arctan(3x) y = arctan ( 3 x ) y = arcsin ( x 2 ) y = \arcsin(x^2) y = arcsin ( x 2 )
d d x arctan ( 3 x ) = 3 1 + 9 x 2 , d d x arcsin ( x 2 ) = 2 x 1 − x 4 \frac{d}{dx}\arctan(3x) = \frac{3}{1 + 9x^2}, \qquad
\frac{d}{dx}\arcsin(x^2) = \frac{2x}{\sqrt{1 - x^4}} d x d arctan ( 3 x ) = 1 + 9 x 2 3 , d x d arcsin ( x 2 ) = 1 − x 4 2 x Each is the outer derivative at the inner function, times the inner derivative.
Example 3 — A logarithm in base 2
Differentiate y = log 2 ( x 2 + 1 ) y = \log_2(x^2 + 1) y = log 2 ( x 2 + 1 )
y ′ = 1 ( x 2 + 1 ) ln 2 ⋅ 2 x = 2 x ( x 2 + 1 ) ln 2 y' = \frac{1}{(x^2 + 1)\ln 2} \cdot 2x = \frac{2x}{(x^2 + 1)\ln 2} y ′ = ( x 2 + 1 ) ln 2 1 ⋅ 2 x = ( x 2 + 1 ) ln 2 2 x Common mistake
Evaluating f ′ f' f ′ In Example 1, g ′ ( 2 ) g'(2) g ′ ( 2 ) 1 f ′ ( 1 ) \tfrac{1}{f'(1)} f ′ ( 1 ) 1 1 f ′ ( 2 ) \tfrac{1}{f'(2)} f ′ ( 2 ) 1 2 2 2 f ′ f' f ′ f f f 2 2 2
Common mistake
Reading arcsin x \arcsin x arcsin x 1 sin x \tfrac{1}{\sin x} s i n x 1 sin − 1 x \sin^{-1} x sin − 1 x csc x \csc x csc x 1 1 − x 2 \tfrac{1}{\sqrt{1 - x^2}} 1 − x 2 1 − csc x cot x -\csc x \cot x − csc x cot x
Let f ( x ) = 2 x + sin x f(x) = 2x + \sin x f ( x ) = 2 x + sin x g = f − 1 g = f^{-1} g = f − 1 g ′ ( 0 ) g'(0) g ′ ( 0 )
Answer
1 3 \tfrac{1}{3} 3 1
Full solution
f ( 0 ) = 0 f(0) = 0 f ( 0 ) = 0 g ( 0 ) = 0 g(0) = 0 g ( 0 ) = 0 f ′ ( x ) = 2 + cos x f'(x) = 2 + \cos x f ′ ( x ) = 2 + cos x f ′ ( 0 ) = 3 f'(0) = 3 f ′ ( 0 ) = 3 g ′ ( 0 ) = 1 3 g'(0) = \tfrac{1}{3} g ′ ( 0 ) = 3 1
Differentiate y = ln ( 5 x ) y = \ln(5x) y = ln ( 5 x )
Answer
y ′ = 1 x y' = \tfrac{1}{x} y ′ = x 1
Full solution
1 5 x ⋅ 5 = 1 x \tfrac{1}{5x} \cdot 5 = \tfrac{1}{x} 5 x 1 ⋅ 5 = x 1 ln ( 5 x ) = ln 5 + ln x \ln(5x) = \ln 5 + \ln x ln ( 5 x ) = ln 5 + ln x
Differentiate y = log 10 x y = \log_{10} x y = log 10 x
Answer
y ′ = 1 x ln 10 y' = \tfrac{1}{x \ln 10} y ′ = x l n 10 1
Full solution
log 10 x = ln x ln 10 \log_{10} x = \tfrac{\ln x}{\ln 10} log 10 x = l n 10 l n x ln 10 \ln 10 ln 10
Differentiate y = arctan ( x 2 ) y = \arctan(x^2) y = arctan ( x 2 )
Answer
y ′ = 2 x 1 + x 4 y' = \tfrac{2x}{1 + x^4} y ′ = 1 + x 4 2 x
Full solution
1 1 + ( x 2 ) 2 ⋅ 2 x \tfrac{1}{1 + (x^2)^2} \cdot 2x 1 + ( x 2 ) 2 1 ⋅ 2 x
Differentiate y = arcsin ( x 2 ) y = \arcsin\!\left(\tfrac{x}{2}\right) y = arcsin ( 2 x )
Answer
y ′ = 1 4 − x 2 y' = \tfrac{1}{\sqrt{4 - x^2}} y ′ = 4 − x 2 1
Full solution
1 1 − x 2 / 4 ⋅ 1 2 = 1 2 ( 4 − x 2 ) / 4 = 1 4 − x 2 \tfrac{1}{\sqrt{1 - x^2/4}} \cdot \tfrac{1}{2} = \tfrac{1}{2\sqrt{(4 - x^2)/4}} = \tfrac{1}{\sqrt{4 - x^2}} 1 − x 2 /4 1 ⋅ 2 1 = 2 ( 4 − x 2 ) /4 1 = 4 − x 2 1
Differentiate y = x arctan x y = x \arctan x y = x arctan x
Answer
y ′ = arctan x + x 1 + x 2 y' = \arctan x + \tfrac{x}{1 + x^2} y ′ = arctan x + 1 + x 2 x
Full solution
Product rule: 1 ⋅ arctan x + x ⋅ 1 1 + x 2 1 \cdot \arctan x + x \cdot \tfrac{1}{1 + x^2} 1 ⋅ arctan x + x ⋅ 1 + x 2 1
Find the tangent line to y = ln x y = \ln x y = ln x x = e x = e x = e
Answer
y = x e y = \tfrac{x}{e} y = e x
Full solution
The point is ( e , 1 ) (e, 1) ( e , 1 ) 1 e \tfrac{1}{e} e 1 y − 1 = 1 e ( x − e ) y - 1 = \tfrac{1}{e}(x - e) y − 1 = e 1 ( x − e ) y = x e y = \tfrac{x}{e} y = e x
Let f ( x ) = e 2 x f(x) = e^{2x} f ( x ) = e 2 x g = f − 1 g = f^{-1} g = f − 1 g ′ ( e 2 ) g'(e^2) g ′ ( e 2 ) g g g
Answer
1 2 e 2 \tfrac{1}{2e^2} 2 e 2 1
Full solution
f ( 1 ) = e 2 f(1) = e^2 f ( 1 ) = e 2 g ( e 2 ) = 1 g(e^2) = 1 g ( e 2 ) = 1 f ′ ( x ) = 2 e 2 x f'(x) = 2e^{2x} f ′ ( x ) = 2 e 2 x f ′ ( 1 ) = 2 e 2 f'(1) = 2e^2 f ′ ( 1 ) = 2 e 2 g ′ ( e 2 ) = 1 2 e 2 g'(e^2) = \tfrac{1}{2e^2} g ′ ( e 2 ) = 2 e 2 1
Check: g ( x ) = 1 2 ln x g(x) = \tfrac{1}{2}\ln x g ( x ) = 2 1 ln x g ′ ( x ) = 1 2 x g'(x) = \tfrac{1}{2x} g ′ ( x ) = 2 x 1 g ′ ( e 2 ) = 1 2 e 2 g'(e^2) = \tfrac{1}{2e^2} g ′ ( e 2 ) = 2 e 2 1
Differentiate y = arccos ( e x ) y = \arccos(e^x) y = arccos ( e x )
Answer
y ′ = − e x 1 − e 2 x y' = -\tfrac{e^x}{\sqrt{1 - e^{2x}}} y ′ = − 1 − e 2 x e x
Full solution
− 1 1 − ( e x ) 2 ⋅ e x -\tfrac{1}{\sqrt{1 - (e^x)^2}} \cdot e^x − 1 − ( e x ) 2 1 ⋅ e x x < 0 x < 0 x < 0 e x < 1 e^x < 1 e x < 1
A student finds g ′ ( 2 ) g'(2) g ′ ( 2 ) g = f − 1 g = f^{-1} g = f − 1 f ( x ) = x 3 + x f(x) = x^3 + x f ( x ) = x 3 + x 1 f ′ ( 2 ) = 1 13 \tfrac{1}{f'(2)} = \tfrac{1}{13} f ′ ( 2 ) 1 = 13 1
Hint
Which input does f f f 2 2 2
Answer
f ′ f' f ′ g ( 2 ) = 1 g(2) = 1 g ( 2 ) = 1 2 2 2 1 4 \tfrac{1}{4} 4 1
Full solution
g ′ ( 2 ) = 1 f ′ ( g ( 2 ) ) g'(2) = \tfrac{1}{f'(g(2))} g ′ ( 2 ) = f ′ ( g ( 2 )) 1 f ( 1 ) = 2 f(1) = 2 f ( 1 ) = 2 g ( 2 ) = 1 g(2) = 1 g ( 2 ) = 1 f ′ ( 1 ) = 3 + 1 = 4 f'(1) = 3 + 1 = 4 f ′ ( 1 ) = 3 + 1 = 4
So g ′ ( 2 ) = 1 4 g'(2) = \tfrac{1}{4} g ′ ( 2 ) = 4 1