3.9 – Inverse Trigonometric Functions

3.9 – Inverse Trigonometric Functions

3.9 – Inverse Trigonometric Functions

𝑦=𝑠𝑖𝑛− 1𝑥2

𝐷𝑥 𝑠𝑖𝑛− 1𝑢=1

√1−𝑢2𝑑𝑢𝑑𝑥

𝑢=𝑥2

𝑦 ′= 1

√1− (𝑥2 )2

𝑦 ′= 2 𝑥√1−𝑥4

𝑦=𝑡𝑎𝑛− 1(3 𝑥)

𝐷𝑥𝑡𝑎𝑛− 1𝑢=1

1+𝑢2𝑑𝑢𝑑𝑥

𝑢=3 𝑥

𝑦 ′= 31+9𝑥2

𝑦 ′= 11+ (3𝑥 )2

𝑦=𝑠𝑒𝑐−1(𝑒¿¿ 2𝑥)¿

𝐷𝑥 𝑠𝑒𝑐− 1𝑢=1

|𝑢|√𝑢2−1𝑑𝑢𝑑𝑥

𝑢=𝑒2𝑥

𝑦 ′= 1

|𝑒2𝑥|√ (𝑒2𝑥 )2−1

𝑦 ′= 2𝑒2𝑥

𝑒2 𝑥√𝑒4 𝑥−1

𝑦 ′= 2√𝑒4 𝑥−1

𝐸𝑥𝑎𝑚𝑝𝑙𝑒𝑠 :

2 𝑥 3 𝑒2𝑥2

3.9 – Inverse Trigonometric Functions

𝑦=(𝑠𝑖𝑛−1 (5 𝑥2 ) )3

2 𝑥

1

√1− (5 𝑥2 )2

𝑦=𝑥2𝑡𝑎𝑛− 1(𝑥4)

𝑦 ′=𝑥2

𝑦 ′=3( 𝑠𝑖𝑛−1 (5 𝑥2 ))2

𝑦 ′=30𝑥 (𝑠𝑖𝑛− 1 (5 𝑥2 ))2

√1−25 𝑥4

10 𝑥

11+(𝑥4 )24 𝑥

3+¿ 𝑡𝑎𝑛− 1(𝑥4)

𝑦 ′= 4 𝑥5

1+𝑥8+2𝑥 𝑡𝑎𝑛−1(𝑥4)

𝐸𝑥𝑎𝑚𝑝𝑙𝑒𝑠 :

𝑦=𝑥2+2𝑑𝑦𝑑𝑥 =2𝑥

𝑎𝑡 𝑥=1 𝑦=3

𝑎𝑡 𝑥=1 𝑑𝑦𝑑𝑥 =2

𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑜𝑓 h𝑡 𝑒 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑙𝑖𝑛𝑒𝑎𝑡 𝑥=1𝑦−3=2(𝑥−1)𝑦=2 𝑥+1

1 1

1

3.11 – Linearization and Differentials

3.11 – Linearization and Differentials

𝑎

𝑑𝑦

∆ 𝑥 𝑜𝑟 𝑑 𝑥

∆ 𝑦

𝑓 (𝑥)𝐿(𝑥)

𝑓 (𝑎)

𝑓 (𝑎 )+𝑑𝑦

𝑥−𝑎

3.11 – Linearization and Differentials

𝑎

𝑑𝑦

∆ 𝑥 𝑜𝑟 𝑑 𝑥

∆ 𝑦

𝑓 (𝑥)𝐿(𝑥)

𝑓 (𝑎)

𝑓 (𝑎 )+𝑑𝑦

𝑥−𝑎

∆ 𝑦∆𝑥 =𝑚𝑠𝑒𝑐

𝑑𝑦𝑑𝑥 = 𝑓 ′ (𝑥 )=𝑚𝑡𝑎𝑛

𝑑𝑦= 𝑓 ′ (𝑥 )𝑑𝑥

𝐴𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑖𝑜𝑛 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛:𝐿(𝑥)

𝐿 (𝑥 )=¿𝑓 (𝑎 )+𝑑𝑦𝐿 (𝑥 )= 𝑓 (𝑎 )+ 𝑓 ′ (𝑎)𝑑𝑥

𝐿 (𝑥 )= 𝑓 (𝑎 )+ 𝑓 ′ (𝑎)(𝑥−𝑎)

𝐿 (𝑥 )≈ 𝑓 (𝑥 )

𝐿𝑖𝑛𝑒𝑎𝑟𝑖𝑧𝑎𝑡𝑖𝑜𝑛𝑜𝑓 𝑓 (𝑥 )

3.11 – Linearization and Differentials

3.11 – Linearization and Differentials

𝐹𝑖𝑛𝑑 h𝑡 𝑒𝑙𝑖𝑛𝑒𝑎𝑟𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 𝑓 (𝑥 )=√𝑥2+16𝑛𝑒𝑎𝑟 𝑥=−3

2 𝑥

𝑓 (𝑥 )=√𝑥2+16¿ (𝑥2+16 )12

𝑓 ′ (𝑥 )= 𝑥√𝑥2+16

𝐿 (𝑥 )= 𝑓 (𝑎 )+ 𝑓 ′ (𝑎)(𝑥−𝑎)

𝐿 (𝑥 )= 𝑓 (−3 )+ 𝑓 ′ (−3 )(𝑥− (−3 ))

𝐿 (𝑥 )=√(−3 )2+16+−3

√ (−3 )2+16(𝑥+3)

𝐿 (𝑥 )=5− 35(𝑥+3 ) 𝐿𝑖𝑛𝑒𝑎𝑟𝑖𝑧𝑎𝑡𝑖𝑜𝑛𝑜𝑓 𝑓 (𝑥 )

𝐸𝑥𝑎𝑚𝑝𝑙𝑒 :

3.11 – Linearization and Differentials𝐹𝑖𝑛𝑑 h𝑡 𝑒𝑙𝑖𝑛𝑒𝑎𝑟 𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 𝑓 (𝑥 )=√𝑥2+16𝑎𝑡 𝑥=−2.95 ,−3.05𝑎𝑛𝑑−4.𝐿 (𝑥 )=5− 35

(𝑥+3 )

𝐿 (−2.95 )=5− 35(−2.95+3 )

𝐿 (−2.95 )=4.97𝑓 (−2.95 )=4.97016

𝐿 (−3.05 )=5− 35(−3.05+3 )

𝐿 (−3.05 )=5.03𝑓 (−3.05 )=5.03016

𝐿 (−4 )=5− 35(−4+3 )

𝐿 (−4 )=5.6

𝑓 (−4 )=5.65685

3.11 – Linearization and Differentials

𝐹𝑖𝑛𝑑 h𝑡 𝑒𝑙𝑖𝑛𝑒𝑎𝑟𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 𝑓 (𝑥 )= tan𝑥𝑛𝑒𝑎𝑟 𝑥=𝜋4

𝑓 (𝑥 )=tan 𝑥𝑓 ′ (𝑥 )=𝑠𝑒𝑐2𝑥𝐿 (𝑥 )= 𝑓 (𝑎 )+ 𝑓 ′ (𝑎)(𝑥−𝑎)

𝐿 (𝑥 )= 𝑓 ( 𝜋4 )+ 𝑓 ′( 𝜋4 )(𝑥− 𝜋4 )𝐿 (𝑥 )= tan 𝜋4

+1

𝑐𝑜𝑠2 𝜋4

𝐿 (𝑥 )=1+2(𝑥− 𝜋4 ) 𝐿𝑖𝑛𝑒𝑎𝑟𝑖𝑧𝑎𝑡𝑖𝑜𝑛𝑜𝑓 𝑓 (𝑥 )

𝐸𝑥𝑎𝑚𝑝𝑙𝑒 :

(𝑥− 𝜋4 )𝐿 (𝑥 )=1

+𝑠𝑒𝑐2 𝜋4

(𝑥− 𝜋4 )

3.11 – Linearization and Differentials

𝐹𝑖𝑛𝑑 h𝑡 𝑒𝑙𝑖𝑛𝑒𝑎𝑟 𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 𝑓 (𝑥 )= tan 𝑥𝑎𝑡 𝑥=2𝜋9 ,

5𝜋18 𝑎𝑛𝑑 𝜋3 .

𝐿( 2𝜋9 )=1+2( 2𝜋9 − 𝜋4 )𝐿( 2𝜋9 )=0.82547𝑓 (2𝜋9 )=0.8391

𝐿 (𝑥 )=1+2(𝑥− 𝜋4 )𝐿( 5𝜋18 )=1+2 (5𝜋18 − 𝜋4 )𝐿( 5𝜋18 )=1.17453𝑓 ( 5𝜋18 )=1.19175

𝐿( 𝜋3 )=1+2(𝜋3 − 𝜋4 )

𝐿( 𝜋3 )=1.5236

𝑓 ( 𝜋3 )=1.73205

3.11 – Linearization and Differentials

𝑈𝑠𝑒𝑙𝑖𝑛𝑒𝑎𝑟𝑖𝑧𝑎𝑡𝑖𝑜𝑛𝑡𝑜𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑒 h𝑡 𝑒𝑣𝑎𝑙𝑢𝑒𝑜𝑓 √99.4𝑓 (𝑥 )=√𝑥𝑓 ′ (𝑥 )=1

2𝑥− 12

h𝐶 𝑜𝑜𝑠𝑒𝑎𝑣𝑎𝑙𝑢𝑒𝑜𝑓 𝑥 𝑛𝑒𝑎𝑟 99.4 .𝑥=100

𝐿 (𝑥 )=√100

𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑒𝑣𝑎𝑙𝑢𝑒𝑜𝑓 √99.4

𝐸𝑥𝑎𝑚𝑝𝑙𝑒 :

+12√100

¿ 𝑥12

¿12√𝑥

𝐿 (𝑥 )= 𝑓 (𝑎 )+ 𝑓 ′ (𝑎)(𝑥−𝑎)

𝐿 (𝑥 )= 𝑓 (100 )+ 𝑓 ′ (100 )(99.4−100)

(99.4−100)

𝐿 (𝑥 )=10+ 120(− .6 )

𝐿 (𝑥 )=10−0.03=9.97

𝑓 (99.4 )=√99.4=9.96995𝑎𝑐𝑡𝑢𝑎𝑙𝑣𝑎𝑙𝑢𝑒𝑜𝑓 √99.4

3.11 – Linearization and Differentials𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑙𝑠 𝑑𝑦

𝑑𝑥 = 𝑓 ′ (𝑥 ) 𝑑𝑦= 𝑓 ′ (𝑥 )𝑑𝑥

𝐸𝑥𝑎𝑚𝑝𝑙𝑒𝑠 :𝑦=𝑒3𝑥𝑑𝑦=¿𝑒3𝑥3𝑑𝑥𝑑𝑦=3𝑒3𝑥𝑑𝑥

𝑦=cos 𝑥2𝑑𝑦=¿−sin 𝑥22 𝑥𝑑𝑥𝑑𝑦=−2𝑥𝑠𝑖𝑛𝑥2𝑑𝑥

𝑥 𝑦 2−4 𝑥32− 𝑦=0

𝑥2 𝑦𝑑𝑦−6 𝑥12 𝑑𝑥−𝑑𝑦=0

2 𝑥𝑦𝑑𝑦−𝑑𝑦=6 𝑥12 𝑑𝑥

𝑑𝑦 (2𝑥𝑦−1 )=6 𝑥12𝑑𝑥

𝑑𝑦=6 𝑥

12

2 𝑥𝑦−1 𝑑𝑥

𝑑𝑦= 6√𝑥2 𝑥𝑦−1 𝑑𝑥

𝑎+𝑑𝑥𝑎

𝑑𝑦

∆ 𝑥 𝑜𝑟 𝑑𝑥

∆ 𝑦

𝑓 (𝑥)𝐿(𝑥)

𝑓 (𝑎)

𝑓 (𝑎+𝑑𝑥)

𝑓 (𝑎 )+𝑑𝑦

𝑥−𝑎

3.11 – Linearization and Differentials𝐴𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑖𝑜𝑛𝐸𝑟𝑟𝑜𝑟 :|∆ 𝑦−𝑑𝑦|

𝑨𝒑𝒑𝒓𝒐𝒙𝒊𝒎𝒂𝒕𝒊𝒐𝒏 𝑬𝒓𝒓𝒐𝒓 :|∆ 𝒚−𝒅𝒚  |

3.11 – Linearization and Differentials𝐴𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑖𝑜𝑛𝐸𝑟𝑟𝑜𝑟 :|∆ 𝑦−𝑑𝑦|𝐸𝑥𝑎𝑚𝑝𝑙𝑒 :

∆ 𝑦= 𝑓 (1+0.1 )− 𝑓 (1)

𝑑𝑦=0.8

𝑓 (𝑥 )=2𝑥2+4 𝑥−3 𝑓𝑜𝑟 𝑥=1𝑎𝑛𝑑𝑑𝑥=0.1

𝑑𝑦=4 𝑥𝑑𝑥+4 𝑑𝑥

∆ 𝑦= 𝑓 (1.1 )− 𝑓 (1)∆ 𝑦=3.82−3∆ 𝑦=0.82

𝑓 (𝑥 )=𝑦=2𝑥2+4 𝑥−3

𝑑𝑦= (4 𝑥+4   )𝑑𝑥𝑑𝑦= (4 (1 )+4   )(0.1)

𝐴𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑖𝑜𝑛𝐸𝑟𝑟𝑜𝑟 :0.2

|∆ 𝑦−𝑑𝑦||0.82−0.8|

3.11 – Linearization and Differentials