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Calculus I, Math 203, Winter 2024

3.6. The Chain Rule

Noushin Sabetghadam

g(x) x f(u) u = g(x)

y = fog(x) = f(g(x)) x

y

0

dy

dx

dy

du

du

dx

y 0 = f 0 (g(x)) · g 0 (x) = f 0 (u) · u 0 (x)

y = [g(x)] n = u n u = g(x)

y

0 = n[g(x)]

n- 1 g

0 (x) = nu

n- 1 u

0

Examples

f(x) = (3x 2

• 4 ) 5

y =

(xe^ + ex)^3

y =

p 2 x - 3

g(t) = ( 3 t t)

10

h(s) =

2s

2

• 3s

1 + s^2

f(✓) =

p ✓ ✓ = ⇡/

h(x) = g(f(x)) h 0 ( 1 ) f( 1 ) = - 2, f 0 ( 1 ) = - 4, f 0 (- 2 ) = 5

g 0 (- 2 ) = 3

h(x) = g(

p x) h 0 ( 4 ) g 0 ( 2 ) = 3

u = u(x)

y = e u y 0 = u 0 e u

y = a u y 0 = u 0 a u a 0

y = u y 0 = u 0

u

u 0

u

y = (^) a u y 0 = u

u a

u

0

u a

y = u y 0 = u 0 u

y = u y

0 = - u

0 u

y = u y 0 = u 0 2 u

y = u y 0 = - u 0 2 u

y = u y 0 = u 0 u u

y = u y 0 = - u 0 u u