How do I use the chain rule to derive certain functions?

In summary, the conversation discusses using the chain rule to derive certain functions, specifically focusing on the outermost functions first before getting to the innermost functions. Examples are provided for functions such as sincos(5x), sin^4(2x) + cos^4(2x), and [ ((3x^2 -2)^5) - 1 ] ^5, with the correct applications of the chain rule and product rule being demonstrated. The conversation also mentions a different answer being obtained for a specific function, highlighting the importance of carefully applying the chain rule.
  • #1
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id like some help deriving certain functions using the chain rule
the way our teacher does it is different from what the textbook says
he derives the outermost functions before getting to the innermost functions, this is where
i get confused =(


for example

f(x) = sincos(5x)
i get
f'(x) = (coscos(5x)) (-sin(5x)) (5)
= -5coscos(5x)sin(5x)

im not sure if this is right

another question:

f(x) = sin^4(2x) + cos^4(2x)
 
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  • #2
The first looks correct... why not have a stab at the second?
 
  • #3
here is what i get

f(x) = sin^4(2x) + cos^4(2x)

f'(x) = [ (4sin^3(2x) (cos^4(2x)) (2) ] + [ (4cos^3(2x) (-sin(2x)) (2) ]

= 8sin^3(2x)cos^4(2x) + -8cos^3(2x)sin(2x)


another question - this one i get a different answer than my friend

f(x) = [ ((3x^2 -2)^5) - 1 ] ^5

i get

f(x) = (3x^10 - 2^5 - 1)^5
= 3x^50 - 2^25 - 1^5

f'(x) = 150x^49

here is what I am working on right now - ill post what i get in a few mins

f(x) = |x^1/3|
f(x) = x^2 cos^2(x^2)
f(x) = [ (2x+5)/(7x-9) ]^2/3
 
  • #4
For the first one in your post you're not applying the chain rule correctly.

d/dx sin(2x)^4 = 4 * sin(2x) ^3 * d/dx (sin2x)

For your second one

(3x^10 - 2^5 - 1)^5 is certainly NOT the same as 3x^50 - 2^25 - 1^5
 
  • #5
for f(x) = sin^4(2x) + cos^4(2x), i am doing it right I am just deriving it wrong? i should derive each part then add together..if so would the answer be8sin^3(2x)cos(2x) - 8cos^3(2x)sin(2x)

for f(x) = [ ((3x^2 -2)^5) - 1 ] ^5
i get

f'(x) = (150x)(3x^2 - 2)^24another one I am sketched on
f(x) = x^2 cos^2(x^2)

using product rule i get

f'(x) = (2x) (cos^2(2x)) + (x^2) (2cos(x^2)(sin(x^2))(2x)
=2xcos^x(x^2) + 4x^3cos(x^2)sin(x^2)
 

1. What is the chain rule in calculus?

The chain rule is a formula used to find the derivative of a composite function. It states that the derivative of a composite function is equal to the derivative of the outer function multiplied by the derivative of the inner function.

2. Why is the chain rule important?

The chain rule is important because it allows us to find the derivative of more complex functions by breaking them down into simpler functions. It is a fundamental tool in calculus and is used in many real-world applications.

3. How do you apply the chain rule?

To apply the chain rule, you must first identify the outer function and the inner function of the composite function. Then, take the derivative of the outer function and plug in the inner function, multiplied by the derivative of the inner function, to get the final derivative.

4. Can you give an example of using the chain rule?

Yes, for example, if we have the function f(x) = (2x^2 + 3)^5, we can break it down into the composite function g(x) = x^5 and h(x) = 2x^2 + 3. Using the chain rule, we can find the derivative of f(x) as f'(x) = 5(2x^2 + 3)^4 * 4x = 20x(2x^2 + 3)^4.

5. Are there any special cases when using the chain rule?

Yes, there are some special cases when using the chain rule, such as when the inner function is a constant or when the inner function is a variable raised to a power. In these cases, the derivative can be simplified further using other rules of differentiation.

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