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Made by: 
Rudra Patel 
Harsh Desai 
Ravi Patel 
Rishabh Patel 
Harshil Raymagiya
We now have a pretty good list of “shortcuts” to find 
derivatives of simple functions. 
Of course, many of the functions that we will encounter 
are not so simple. What is needed is a way to combine 
derivative rules to evaluate more complicated functions. 
®
Consider a simple composite function: 
y = 6x -10 
y = 2( 3x -5) 
If u = 3x -5 
then y = 2u 
y = 6x -10 y = 2u u = 3x - 5 
= 2 dy 
dy 6 
dx 
du 
= du 3 
dx 
= 
6 = 2×3 
dy = dy × 
du 
dx du dx 
®
and another: 
y = 5u - 2 
where u = 3t 
then y = 5( 3t ) - 2 
u = 3t 
dy 15 
dt 
= dy 5 
du 
= du 3 
dt 
= 
15 = 5×3 
dy = dy × 
du 
dt du dt 
y = 5( 3t ) - 2 
y =15t - 2 
y = 5u - 2 
®
and one more: 
y = 9x2 + 6x +1 
( ) 2 y = 3x +1 
If u = 3x +1 
u = 3x +1 
dy 18x 6 
dx 
y = u2 
= + dy 2u 
du 
= du 3 
dx 
= 
dy = dy × 
du 
dx du dx 
then y = u2 
y = 9x2 + 6x +1 
dy 2( 3x 1) 
du 
= + 
dy = 6x + 
2 
du 
18x + 6 = ( 6x + 2) ×3 
This pattern is called 
the chain rule. 
®
dy dy du 
dx du dx 
Chain Rule: = × 
f o g y = f ( u) u = g ( x) 
If is the composite of and , 
then: 
o ¢ = ¢ × ¢ 
( ) at u g( x) at x f g f g = 
example: f ( x) = sin x g ( x) = x2 - 4 Find: ( f o g )¢ at x = 2 
f ¢( x) = cos x g¢( x) = 2x g ( 2) = 4 - 4 = 0 
f ¢( 0) × g¢( 2) 
cos ( 0) ×( 2×2) 
1× 4 = 4 
®
We could also do it this way: 
f ( g ( x) ) = sin ( x2 - 4) 
y = sin ( x2 - 4) 
y = sin u u = x2 - 4 
dy = cosu 
du = 
2x 
du 
dx 
dy = dy × 
du 
dx du dx 
dy = cosu × 
2x 
dx 
dy = cos ( x2 - 4) × 
2x 
dx 
dy = cos ( 22 - 4) × 2 × 
2 
dx 
dy = cos( 0) × 
4 
dx 
dy = 
4 
dx 
®
Here is a faster way to find the derivative: 
y = sin ( x2 - 4) 
y ¢ = cos( x2 - 4) ´ d ( x2 - 
4) 
dx 
y¢ = cos ( x2 - 4) ´2x 
Differentiate the outside function... 
…then the inside function 
At x = 2, y¢ = 4 
®
Another example: 
d cos2 ( 3x) 
dx 
d éë cos ( 3x 
) 2 dx 
ùû 
2 cos ( 3x) d cos ( 3x) 
dx 
éë ùû × 
derivative of the 
outside function 
derivative of the 
inside function 
It looks like we need to 
use the chain rule again! 
®
Another example: 
d cos2 ( 3x) 
dx 
d éë cos ( 3x 
) 2 dx 
ùû 
2 cos ( 3x) d cos ( 3x) 
dx 
éë ùû × 
2cos ( 3x) sin ( 3x) d ( 3x) 
dx 
×- × 
-2cos ( 3x) ×sin ( 3x) ×3 
-6cos ( 3x) sin ( 3x) 
The chain rule can be used 
more than once. 
(That’s what makes the 
“chain” in the “chain rule”!) 
®
Derivative formulas include the chain rule! 
d un = nun - 1 du 
d sin u = 
cosu du 
dx dx 
dx dx 
d cosu sin u du 
dx dx 
= - d tan u = 
sec2 u du 
dx dx 
etcetera… 
The formulas on the memorization sheet are written with 
instead of . Don’t forget to include the term! 
u¢ 
du u¢ 
dx 
®
The most common mistake on the chapter 3 test is to 
forget to use the chain rule. 
Every derivative problem could be thought of as a chain-rule 
problem: 
d x2 
dx 
2x d x 
dx 
= = 2x ×1 = 2x 
derivative of 
outside function 
derivative of 
inside function 
The derivative of x is one. 
®
The chain rule enables us to find the slope of 
parametrically defined curves: 
dy = dy × 
dx 
dt dx dt 
dy 
dt dy 
dx = 
dx 
dt 
Divide both sides by 
dx 
The slope of a pardatmetrized 
curve is given by: 
dy 
dy dt 
dx dx 
dt 
= 
®
Example: x = 3cos t y = 2sin t 
These are the equations for 
an ellipse. 
dx 3sin t 
dt 
= - dy 2cos t 
dt 
dy t 
dx t 
= 2cos 
= 
- 
3sin 
2 cot 
3 
= - t
p

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Chain Rule

  • 1. Made by: Rudra Patel Harsh Desai Ravi Patel Rishabh Patel Harshil Raymagiya
  • 2. We now have a pretty good list of “shortcuts” to find derivatives of simple functions. Of course, many of the functions that we will encounter are not so simple. What is needed is a way to combine derivative rules to evaluate more complicated functions. ®
  • 3. Consider a simple composite function: y = 6x -10 y = 2( 3x -5) If u = 3x -5 then y = 2u y = 6x -10 y = 2u u = 3x - 5 = 2 dy dy 6 dx du = du 3 dx = 6 = 2×3 dy = dy × du dx du dx ®
  • 4. and another: y = 5u - 2 where u = 3t then y = 5( 3t ) - 2 u = 3t dy 15 dt = dy 5 du = du 3 dt = 15 = 5×3 dy = dy × du dt du dt y = 5( 3t ) - 2 y =15t - 2 y = 5u - 2 ®
  • 5. and one more: y = 9x2 + 6x +1 ( ) 2 y = 3x +1 If u = 3x +1 u = 3x +1 dy 18x 6 dx y = u2 = + dy 2u du = du 3 dx = dy = dy × du dx du dx then y = u2 y = 9x2 + 6x +1 dy 2( 3x 1) du = + dy = 6x + 2 du 18x + 6 = ( 6x + 2) ×3 This pattern is called the chain rule. ®
  • 6. dy dy du dx du dx Chain Rule: = × f o g y = f ( u) u = g ( x) If is the composite of and , then: o ¢ = ¢ × ¢ ( ) at u g( x) at x f g f g = example: f ( x) = sin x g ( x) = x2 - 4 Find: ( f o g )¢ at x = 2 f ¢( x) = cos x g¢( x) = 2x g ( 2) = 4 - 4 = 0 f ¢( 0) × g¢( 2) cos ( 0) ×( 2×2) 1× 4 = 4 ®
  • 7. We could also do it this way: f ( g ( x) ) = sin ( x2 - 4) y = sin ( x2 - 4) y = sin u u = x2 - 4 dy = cosu du = 2x du dx dy = dy × du dx du dx dy = cosu × 2x dx dy = cos ( x2 - 4) × 2x dx dy = cos ( 22 - 4) × 2 × 2 dx dy = cos( 0) × 4 dx dy = 4 dx ®
  • 8. Here is a faster way to find the derivative: y = sin ( x2 - 4) y ¢ = cos( x2 - 4) ´ d ( x2 - 4) dx y¢ = cos ( x2 - 4) ´2x Differentiate the outside function... …then the inside function At x = 2, y¢ = 4 ®
  • 9. Another example: d cos2 ( 3x) dx d éë cos ( 3x ) 2 dx ùû 2 cos ( 3x) d cos ( 3x) dx éë ùû × derivative of the outside function derivative of the inside function It looks like we need to use the chain rule again! ®
  • 10. Another example: d cos2 ( 3x) dx d éë cos ( 3x ) 2 dx ùû 2 cos ( 3x) d cos ( 3x) dx éë ùû × 2cos ( 3x) sin ( 3x) d ( 3x) dx ×- × -2cos ( 3x) ×sin ( 3x) ×3 -6cos ( 3x) sin ( 3x) The chain rule can be used more than once. (That’s what makes the “chain” in the “chain rule”!) ®
  • 11. Derivative formulas include the chain rule! d un = nun - 1 du d sin u = cosu du dx dx dx dx d cosu sin u du dx dx = - d tan u = sec2 u du dx dx etcetera… The formulas on the memorization sheet are written with instead of . Don’t forget to include the term! u¢ du u¢ dx ®
  • 12. The most common mistake on the chapter 3 test is to forget to use the chain rule. Every derivative problem could be thought of as a chain-rule problem: d x2 dx 2x d x dx = = 2x ×1 = 2x derivative of outside function derivative of inside function The derivative of x is one. ®
  • 13. The chain rule enables us to find the slope of parametrically defined curves: dy = dy × dx dt dx dt dy dt dy dx = dx dt Divide both sides by dx The slope of a pardatmetrized curve is given by: dy dy dt dx dx dt = ®
  • 14. Example: x = 3cos t y = 2sin t These are the equations for an ellipse. dx 3sin t dt = - dy 2cos t dt dy t dx t = 2cos = - 3sin 2 cot 3 = - t
  • 15. p