Understand the idea An inverse derivative uses the slope of the original function at the corresponding input, not at the same numeric argument. Require local invertibility and a nonzero original derivative. The derivatives of inverse trigonometric functions can be derived implicitly; apply the chain rule to their arguments. The principal branches determine the relevant domains.
( f − 1 ) ′ ( b ) = 1 f ′ ( a ) where f ( a ) = b ; ( arcsin x ) ′ = 1 1 − x 2 , ( arccos x ) ′ = − 1 1 − x 2 , ( arctan x ) ′ = 1 1 + x 2 (f^{-1})'(b)=\frac1{f'(a)}\quad\text{where }f(a)=b;\qquad(\arcsin x)'=\frac1{\sqrt{1-x^2}},\quad(\arccos x)'=-\frac1{\sqrt{1-x^2}},\quad(\arctan x)'=\frac1{1+x^2} ( f − 1 ) ′ ( b ) = f ′ ( a ) 1 where f ( a ) = b ; ( arcsin x ) ′ = 1 − x 2 1 , ( arccos x ) ′ = − 1 − x 2 1 , ( arctan x ) ′ = 1 + x 2 1 Worked example If f ( 2 ) = 5 \displaystyle f(2)=5 f ( 2 ) = 5 f ′ ( 2 ) = 3 \displaystyle f'(2)=3 f ′ ( 2 ) = 3 ( f − 1 ) ′ ( 5 ) \displaystyle (f^{-1})'(5) ( f − 1 ) ′ ( 5 )
The inverse sends
5 \displaystyle 5 5 to
2 \displaystyle 2 2 . Reciprocal slopes at corresponding points give
( f − 1 ) ′ ( 5 ) = 1 / f ′ ( 2 ) = 1 / 3 \displaystyle (f^{-1})'(5)=1/f'(2)=1/3 ( f − 1 ) ′ ( 5 ) = 1/ f ′ ( 2 ) = 1/3 .
Check this before moving on f − 1 \displaystyle f^{-1} f − 1 1 / f \displaystyle 1/f 1/ f
Try it yourself Differentiate y = arctan ( 3 x ) \displaystyle y=\arctan(3x) y = arctan ( 3 x )
Show solution and reasoning The chain rule gives y ′ = 3 / ( 1 + 9 x 2 ) \displaystyle y'=3/(1+9x^2) y ′ = 3/ ( 1 + 9 x 2 )
Work through AP Calculus AB with a tutor Detailed notes are discussed during the course with worked examples, guided practice, and feedback on mathematical reasoning.
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