Microsoft Math 3.0: Solving Complex Equations

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Homework Help Overview

The discussion revolves around solving complex equations involving composite functions and their derivatives, particularly using the chain rule in calculus. Participants are exploring various approaches to differentiate functions that are nested within each other, as well as addressing the challenges posed by multiple parentheses in expressions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to apply the chain rule to differentiate composite functions, with some expressing confusion over the complexity of the expressions. There are discussions about breaking down functions into simpler parts and applying differentiation step by step. Some participants question the correctness of their approaches and seek clarification on the differentiation process.

Discussion Status

The discussion is ongoing, with several participants providing hints and partial solutions. There is a mix of attempts to solve specific problems and requests for guidance on general principles of differentiation. Some participants have offered insights into how to approach the differentiation of nested functions, while others are still grappling with the initial steps.

Contextual Notes

Participants note constraints such as the inability to use calculators or software for complex calculations, which adds to the challenge of solving the problems presented. There are also references to specific steps being incorrect, indicating a need for careful verification of the differentiation process.

ARYT
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Homework Statement



Homework Equations



Answers by Microsoft Math 3.0

The Attempt at a Solution



This is confusing. Too many parentheses. We used to solve a composite of two or three functions.
 
Last edited by a moderator:
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If f(x) = h(g(x)) then f'(x) = h'(g(x)) × g'(x).
In words: differentiate the ‘outside’ function, and then multiply by the derivative of the
‘inside’ function.

Let's take an example,
y= sin (x2)
dy/dx= d(sin x2)/dx *d(x2)/dx

Treat all the parentheses in a similar manner.
 
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I know, I could solve the third one myself, but the first one :(
 
Break up the expression as:
z(x) = (1+v(x))5
v(x) = (2-u(x))3
u(x) = (6+7x2)9

finally you have y = 10*z(x)

now dy/dx = dy/dz *dz/dx
 
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(fog(x) )'=g' (x) f' (g(x) )

we have this general rule for two functions only. Give me sth for n functions.
 
ARYT said:
(fog(x) )'=g' (x) f' (g(x) )

we have this general rule for two functions only. Give me sth for n functions.

shramana has given you a pretty good hint. How about you try to follow it? Post your work, and we will point out any errors.
 
Now let's say
g(x)= g(h(x))
so you'll have g'(x)=h'(x)*g'(h(x)) [using f'(x)=g' (x) f' (g(x) )]

substitute the value of g(x) in f'(x) and so on...
 
Do it step by step: If y= 10u5, the dy/dx= 50u4 du/dx.

And u= 1+v3 so du/dx= 3v^2 dv/dx.

v= ...?
 
OK. Here I've tried to solve the first one.
 
Last edited by a moderator:
  • #10
Although it's too long. I won't be able to do it without a software (for multiplication and things like that). Also, We can't use calculator.
 
  • #11
And the second one which compare to the answer given by the Microsoft Math is wrong:
 
Last edited by a moderator:
  • #12
ARYT said:
Although it's too long. I won't be able to do it without a software (for multiplication and things like that). Also, We can't use calculator.

Not at all. You just have to differentiate the functions in the parentheses as you move inwards.

f'(x)=10*z'(x)
z'(x)=5*(1+v(x))4*v'(x)
v'(x)=3*(1-u(x))2*(-u'(x))
u'(x)=9*7*d(x4)/dx

Now substitute the values of u'(x), v'(x),z'(x) and u(x), v(x) in f'(x).
 
  • #13
ARYT said:
And the second one which compare to the answer given by the Microsoft Math is wrong:

Step 2 is wrong.
v'(x) = d(ln(ln sec x))/dx
 

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