Write the functions in the form of y=f(u) and u = g(x), then find dy/dx as a function of
using chain rule.
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1)y = (2x + 1)5
2)y = sec (tan x)
3)y = 3x2 - 4x + 6
plz reply plz
Answers
Step-by-step explanation:
Learning Objectives
State the chain rule for the composition of two functions.
Apply the chain rule together with the power rule.
Apply the chain rule and the product/quotient rules correctly in combination when both are necessary.
Recognize the chain rule for a composition of three or more functions.
Describe the proof of the chain rule.
We have seen the techniques for differentiating basic functions (, etc.) as well as sums, differences, products, quotients, and constant multiples of these functions. However, these techniques do not allow us to differentiate compositions of functions, such as  or . In this section, we study the rule for finding the derivative of the composition of two or more functions.
Deriving the Chain Rule
When we have a function that is a composition of two or more functions, we could use all of the techniques we have already learned to differentiate it. However, using all of those techniques to break down a function into simpler parts that we are able to differentiate can get cumbersome. Instead, we use the chain rule, which states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.
To put this rule into context, let’s take a look at an example: . We can think of the derivative of this function with respect to  as the rate of change of  relative to the change in . Consequently, we want to know how  changes as  changes. We can think of this event as a chain reaction: As  changes,  changes, which leads to a change in . This chain reaction gives us hints as to what is involved in computing the derivative of . First of all, a change in  forcing a change in  suggests that somehow the derivative of  is involved. In addition, the change in  forcing a change in  suggests that the derivative of  with respect to , where , is also part of the final derivative.
We can take a more formal look at the derivative of  by setting up the limit that would give us the derivative at a specific value  in the domain of .
.
This expression does not seem particularly helpful; however, we can modify it by multiplying and dividing by the expression  to obtain
.
From the definition of the derivative, we can see that the second factor is the derivative of  at . That is,
.
However, it might be a little more challenging to recognize that the first term is also a derivative. We can see this by letting  and observing that as :

Thus, .