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Ideas in this lesson Idea 1 of 4
Derivative of arcsine d ( arcsin x ) d x = 1 1 − x 2 \frac{d(\arcsin x)}{dx} = \frac{1}{\sqrt{1 - x^{2}}} d x d ( a r c s i n x ) = 1 − x 2 1 − 1 < x < 1 -1 < x < 1 − 1 < x < 1
Worked example 1 Let f ( x ) = 2 arcsin x f(x) = 2 \arcsin x f ( x ) = 2 arcsin x f ′ ( 0 ) f'(0) f ′ ( 0 ) f ′ ( x ) = 2 1 − x 2 f'(x) = \dfrac{2}{\sqrt{1 - x^{2}}} f ′ ( x ) = 1 − x 2 2 f ′ ( 0 ) = 2 1 − ( 0 ) 2 f'(0) = \dfrac{2}{\sqrt{1 - (0)^{2}}} f ′ ( 0 ) = 1 − ( 0 ) 2 2 f ′ ( 0 ) = 2 f'(0) = 2 f ′ ( 0 ) = 2 Answer: 2 2 2
Worked example 2 Let f ( x ) = 3 arcsin x f(x) = 3 \arcsin x f ( x ) = 3 arcsin x f ′ ( 0.5 ) f'(0.5) f ′ ( 0.5 ) f ′ ( x ) = 3 1 − x 2 f'(x) = \dfrac{3}{\sqrt{1 - x^{2}}} f ′ ( x ) = 1 − x 2 3 f ′ ( 0.5 ) = 3 1 − ( 0.5 ) 2 f'(0.5) = \dfrac{3}{\sqrt{1 - (0.5)^{2}}} f ′ ( 0.5 ) = 1 − ( 0.5 ) 2 3 f ′ ( 0.5 ) = 6 3 ≈ 3.46 f'(0.5) = \dfrac{6}{\sqrt{3}} \approx 3.46 f ′ ( 0.5 ) = 3 6 ≈ 3.46 Answer: 6 3 ≈ 3.46 \dfrac{6}{\sqrt{3}} \approx 3.46 3 6 ≈ 3.46
Idea 2 of 4
Derivative of arccosine d ( arccos x ) d x = − 1 1 − x 2 \frac{d(\arccos x)}{dx} = -\frac{1}{\sqrt{1 - x^{2}}} d x d ( a r c c o s x ) = − 1 − x 2 1
Worked example 1 Let f ( x ) = 2 arccos x f(x) = 2 \arccos x f ( x ) = 2 arccos x f ′ ( 0 ) f'(0) f ′ ( 0 ) f ′ ( x ) = − 2 1 − x 2 f'(x) = -\dfrac{2}{\sqrt{1 - x^{2}}} f ′ ( x ) = − 1 − x 2 2 f ′ ( 0 ) = − 2 1 − ( 0 ) 2 f'(0) = -\dfrac{2}{\sqrt{1 - (0)^{2}}} f ′ ( 0 ) = − 1 − ( 0 ) 2 2 f ′ ( 0 ) = − 2 f'(0) = -2 f ′ ( 0 ) = − 2 Answer: − 2 -2 − 2
Worked example 2 Let f ( x ) = 3 arccos x f(x) = 3 \arccos x f ( x ) = 3 arccos x f ′ ( 0.5 ) f'(0.5) f ′ ( 0.5 ) f ′ ( x ) = − 3 1 − x 2 f'(x) = -\dfrac{3}{\sqrt{1 - x^{2}}} f ′ ( x ) = − 1 − x 2 3 f ′ ( 0.5 ) = − 3 1 − ( 0.5 ) 2 f'(0.5) = -\dfrac{3}{\sqrt{1 - (0.5)^{2}}} f ′ ( 0.5 ) = − 1 − ( 0.5 ) 2 3 f ′ ( 0.5 ) = − 6 3 ≈ − 3.46 f'(0.5) = -\dfrac{6}{\sqrt{3}} \approx -3.46 f ′ ( 0.5 ) = − 3 6 ≈ − 3.46 Answer: − 6 3 ≈ − 3.46 -\dfrac{6}{\sqrt{3}} \approx -3.46 − 3 6 ≈ − 3.46
Idea 3 of 4
Derivative of arctangent d ( arctan x ) d x = 1 1 + x 2 \frac{d(\arctan x)}{dx} = \frac{1}{1 + x^{2}} d x d ( a r c t a n x ) = 1 + x 2 1 x x x
Worked example 1 Let f ( x ) = 2 arctan x f(x) = 2 \arctan x f ( x ) = 2 arctan x f ′ ( 1 ) f'(1) f ′ ( 1 ) f ′ ( x ) = 2 1 + x 2 f'(x) = \dfrac{2}{1 + x^{2}} f ′ ( x ) = 1 + x 2 2 f ′ ( 1 ) = 2 1 + 1 2 f'(1) = \dfrac{2}{1 + 1^{2}} f ′ ( 1 ) = 1 + 1 2 2 f ′ ( 1 ) = 1 f'(1) = 1 f ′ ( 1 ) = 1 Answer: 1 1 1
Worked example 2 Let f ( x ) = 3 arctan x f(x) = 3 \arctan x f ( x ) = 3 arctan x f ′ ( 3 ) f'(3) f ′ ( 3 ) f ′ ( x ) = 3 1 + x 2 f'(x) = \dfrac{3}{1 + x^{2}} f ′ ( x ) = 1 + x 2 3 f ′ ( 3 ) = 3 1 + 3 2 f'(3) = \dfrac{3}{1 + 3^{2}} f ′ ( 3 ) = 1 + 3 2 3 f ′ ( 3 ) = 3 10 ≈ 0.3 f'(3) = \dfrac{3}{10} \approx 0.3 f ′ ( 3 ) = 10 3 ≈ 0.3 Answer: 3 10 ≈ 0.3 \dfrac{3}{10} \approx 0.3 10 3 ≈ 0.3
Idea 4 of 4
Scaled inverse-trigonometric inputs An inverse-trig composition needs the input derivative as well as the standard denominator.
Worked example 1 Find f ′ ( 0 ) f'(0) f ′ ( 0 ) f ( x ) = arcsin ( x 2 ) f(x) = \arcsin (\dfrac{x}{2}) f ( x ) = arcsin ( 2 x ) f ′ ( x ) = 1 4 − x 2 f'(x) = \dfrac{1}{\sqrt{4 - x^{2}}} f ′ ( x ) = 4 − x 2 1 f ′ ( 0 ) = 1 2 ≈ 0.5 f'(0) = \dfrac{1}{2} \approx 0.5 f ′ ( 0 ) = 2 1 ≈ 0.5 Answer: 1 2 ≈ 0.5 \dfrac{1}{2} \approx 0.5 2 1 ≈ 0.5
Worked example 2 Find f ′ ( 0 ) f'(0) f ′ ( 0 ) f ( x ) = arctan ( 3 x ) f(x) = \arctan (3x) f ( x ) = arctan ( 3 x ) f ′ ( x ) = 3 1 + 9 x 2 f'(x) = \dfrac{3}{1 + 9x^{2}} f ′ ( x ) = 1 + 9 x 2 3 f ′ ( 0 ) = 3 f'(0) = 3 f ′ ( 0 ) = 3 Answer: 3 3 3
Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §5.6 , p 5.6, p 5.6 , p 366 366 366