The Chain Rule!

In words, to find the derivative of f[v(x)], find the derivative of f(x), replace each x with v(x) and then multiply the result by the derivative of v(x).
 

Solution:

v(x) = 14x2 + 1	f(v) = v1/2
v'(x) = 28x

Basically, we are making a substitution.  We are replacing 14x² + 1 with the function v(x) and then taking the derivative of the substitution.  Now apply the formula.





Solution:

v(x) = (3 + 5x)	f(v) = v3
v'(x) = 5	f''(v) = 3v2 or = 3(3+5x)2

f '(x) = 3(3 + 5x)^{2 .} (5) = 15(3+5x)²
= 15(9 + 30x + 25x²) = 135 + 450x + 375x²





Solution:

v(x) = (3 + 5x)	f(v) = v-3/4
v'(x) = 5





4)  f(x) = 4x(3x + 5)⁵  Find f '(x)
Solution:

This involves both the product rule and chain rule.  Use the chain rule to find the derivative of the second function, then apply the product rule!

v(x) = (3x + 5)	f(v) = v5
v'(x) = 3	f '(v) = 5v4 or 5(3x + 5)4

Using the chain rule gives us f '(x) = 15(3x + 5)⁴
Now use the product rule!

r(x) = 4x	s(x) = (3x + 5)5
r'(x) = 4	s'(x) = 15(3x + 5)4

f '(x) = (4x)(15)(3x + 5)⁴ + 4(3x + 5)⁵
= 60x(3x + 5)⁴ + 4(3x + 5)⁵
= 4(3x + 5)⁴(15x + 3x + 5)
= 4(3x + 5)⁴(18x + 5)



   Find f ' (x).
Solution:

This involves using the chain rule in the quotient rule.  First apply the chain rule to find the derivative of the numerator and then apply the product rule!

v(x) = (3x + 2)	f(v) = v7
v'(x) = 3	f '(v) = 7v6 or 7(3x + 2)6

f '(x) = (3)(7)(3x + 2)⁶ = 21(3x + 2)^{6
Now use the quotient rule!!}

r(x) = (3x + 2)7	s(x) = (x - 1)
r'(x) = 21(3x + 2)6	s'(x) = 1








 
That's it for the chain rule!!  On to the derivative of the exponential and log functions.