3.7 – Implicit DifferentiationAn Implicit function is one where the variable “y” can not be easily solved for in terms of only “x”.Examples:

4 𝑥2=2 𝑦3+4 𝑦

𝑥=√ tan 𝑦3 𝑥2𝑦 2=4 𝑥2−4 𝑥𝑦

𝑥2sin 𝑥+𝑦2 cos 𝑦=¿1¿

3.7 – Implicit DifferentiationDifferentiating an implicit function will create the derivative () needed to calculate the slope of any tangent on the curve or the rate of change at any value of the independent variable.Find

6 𝑥 𝑑𝑥𝑑𝑥 +¿

3 𝑥2+𝑦2=14

2 𝑦 𝑑𝑦𝑑𝑥 =¿0

6 𝑥+2 𝑦 𝑑𝑦𝑑𝑥=0

2 𝑦 𝑑𝑦𝑑𝑥 =−6 x

𝑑𝑦𝑑𝑥 =− 6 𝑥2 𝑦

𝑑𝑦𝑑𝑥 =− 3 𝑥𝑦

3.7 – Implicit DifferentiationFind

𝑥35 𝑦4 𝑑𝑦𝑑𝑥 +¿

𝑥3 𝑦5+3 𝑥=8 𝑦3+13 𝑑𝑥𝑑𝑥=¿3 𝑥2 𝑑𝑥𝑑𝑥 𝑦5+¿ 24 𝑦2 𝑑𝑦𝑑𝑥 +0

5 𝑥3 𝑦4 𝑑𝑦𝑑𝑥 +3 𝑥2 𝑦5+3=24 𝑦2 𝑑𝑦𝑑𝑥

5 𝑥3 𝑦4 𝑑𝑦𝑑𝑥 −24 𝑦2 𝑑𝑦𝑑𝑥=−3 𝑥2 𝑦5−3

𝑑𝑦𝑑𝑥 (5 𝑥3 𝑦4−24 𝑦2 )=−3𝑥2 𝑦5−3

𝑑𝑦𝑑𝑥=

−3 𝑥2 𝑦5−3  5 𝑥3 𝑦4−24 𝑦 2

3.7 – Implicit Differentiation

3.7 – Implicit DifferentiationFind the equation of the tangent and normal lines for the following implicit function at the given point.

3 𝑥2+𝑦2=14

𝑦−√2=−3√2 (𝑥−2 )

(2 ,√2 )𝑑𝑦𝑑𝑥=− 3 𝑥𝑦

𝑚𝑡𝑎𝑛=𝑑𝑦𝑑𝑥=− 3 (2 )

√2

𝑚𝑡𝑎𝑛=𝑑𝑦𝑑𝑥=− 6

√2=−3√2

𝑚𝑡𝑎𝑛=−6√2

=−3 √2

𝑚𝑛𝑜𝑟𝑚𝑎𝑙=√26

𝑦−√2=√26

(𝑥−2 )

3.8 – Derivatives of Inverse Functions and Logarithms

𝑦=𝑙𝑛 𝑥3𝑑𝑦𝑑𝑥 =

1𝑥3

3.8 – Derivatives of Inverse Functions and Logarithms

Find

(3 𝑥2 )

𝑦=cos (𝑙𝑛 (4 𝑥3 ))𝑑𝑦𝑑𝑥 =−sin ( 𝑙𝑛 (4 𝑥3 )) (12 𝑥2)

𝑑𝑦𝑑𝑥 =−

3sin (𝑙𝑛 (4 𝑥3 ) )𝑥

𝑑𝑦𝑑𝑥=

3𝑥

𝑦=𝑙𝑛 𝑥3𝑑𝑦𝑑𝑥=3 1𝑥𝑑𝑦𝑑𝑥 =

3𝑥

¿3 𝑙𝑛𝑥

14 𝑥3

𝑦=44𝑥4

𝑑𝑦𝑑𝑥 =44 𝑥

4

3.8 – Derivatives of Inverse Functions and Logarithms

Find

(𝑙𝑛 4)(16 𝑥3 )

𝑦=3𝑐𝑜𝑠 𝑥4

𝑑𝑦𝑑𝑥=3𝑐𝑜𝑠 𝑥

4

(𝑙𝑛3)(−𝑠𝑖𝑛𝑥4 )(4 𝑥3 )

𝑑𝑦𝑑𝑥 =−4 𝑥33𝑐𝑜𝑠𝑥

4

𝑠𝑖𝑛𝑥4 𝑙𝑛3

𝑦=𝑙𝑜𝑔63 𝑥2

𝑑𝑦𝑑𝑥 =

13 𝑥2

3.8 – Derivatives of Inverse Functions and Logarithms

Find

1𝑙𝑛66 𝑥

𝑦=𝑙𝑜𝑔3 (3𝑥5+5 )5

𝑑𝑦𝑑𝑥 =

1(3 𝑥5+5 )5

1𝑙𝑛35 (3 𝑥5+5 )4(15 𝑥4 )

𝑑𝑦𝑑𝑥=

2𝑥𝑙𝑛6

𝑦=𝑙𝑜𝑔24 𝑥5

𝑑𝑦𝑑𝑥 =

14 𝑥5

1𝑙𝑛220 𝑥

4

𝑑𝑦𝑑𝑥=

5𝑥𝑙𝑛2

𝑑𝑦𝑑𝑥 =

75𝑥4

(3 𝑥5+5 ) 𝑙𝑛3

3.8 – Derivatives of Inverse Functions and Logarithms

This technique is useful in cases where it is easier to differentiate the logarithm of a function rather than the function itself. 

Logarithmic Differentiation

𝑦=2 𝑥2𝑥 𝑦=𝑥sin 𝑥 𝑦=(1−𝑥 )2 (𝑥+1 )4

𝑦=2 𝑥2𝑥

𝑙𝑛𝑦=ln (2 𝑥2𝑥)𝑙𝑛𝑦=ln 2+ln (𝑥2 𝑥)𝑙𝑛𝑦=ln 2+2 xln 𝑥1𝑦𝑑𝑦𝑑𝑥=0+2 x 1𝑥 +2 ln 𝑥

1𝑦𝑑𝑦𝑑𝑥=2+2 ln𝑥

𝑑𝑦𝑑𝑥= y ¿

𝑑𝑦𝑑𝑥 =2𝑥2𝑥 ¿

𝑑𝑦𝑑𝑥=4 𝑥2𝑥 ¿

3.8 – Derivatives of Inverse Functions and LogarithmsLogarithmic Differentiation

𝑦=𝑥sin 𝑥𝑙𝑛𝑦= ln (𝑥sin 𝑥)𝑙𝑛𝑦=sin x ln 𝑥1𝑦𝑑𝑦𝑑𝑥=sin x 1𝑥 +cos 𝑥 ln 𝑥

𝑑𝑦𝑑𝑥 = y ¿

𝑑𝑦𝑑𝑥 =𝑥sin 𝑥 ¿

3.8 – Derivatives of Inverse Functions and LogarithmsLogarithmic Differentiation

𝑦=(1−𝑥 )2 (𝑥+1 )4

𝑙𝑛𝑦= ln ((1−𝑥 )2 (𝑥+1 )4)

𝑙𝑛𝑦=2 ln (1−𝑥 )+4 ln (𝑥+1)𝑙𝑛𝑦= ln (1−𝑥 )2+𝑙𝑛 (𝑥+1 )4

1𝑦𝑑𝑦𝑑𝑥=2 1

1−𝑥 (−1)+4 1𝑥+1

𝑑𝑦𝑑𝑥 =𝑦 ( −21−𝑥 + 4

𝑥+1 )𝑑𝑦𝑑𝑥 =(1−𝑥 )2 (𝑥+1 )4 ( −21−𝑥 + 4

𝑥+1 )