Calculus - find dy/dt thank you

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    Calculus Thank you
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Homework Help Overview

The discussion revolves around a calculus problem involving the differentiation of a function expressed as dy/dt = x^2 tan(x) / sec(x). The original poster expresses difficulty in matching the book's answer and seeks assistance in understanding the steps involved.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts various differentiation techniques, including the quotient and product rules, but struggles to arrive at the correct answer. Another participant clarifies the expression and suggests a simplification that leads to a more straightforward differentiation.

Discussion Status

The conversation is ongoing, with one participant providing a potential simplification of the original expression. There is no explicit consensus on the correct method yet, but a direction for further exploration has been suggested.

Contextual Notes

The original poster is seeking help within the constraints of homework guidelines, indicating a desire for guidance rather than complete solutions.

laker_gurl3
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calculus -.. find dy/dt...thank you..:(

can someone please help me out with this question...i did it all sorts of way,quotient rule, product rule..etc.. but i can never match the answer in the book..i just want to know if u did..

find dy/dt = x^2 Tanx / sec x

The answer in the book was
x(xcosx + 2sinx)

lemmi know how you did it with a couple steps! thanks soo mcuh
 
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Do you mean find

[tex]\frac{d}{dx} \frac{x^2 \tan{x}}{\sec{x}}[/tex]

?

If so, here:

[tex]\frac{x^2 \tan{x}}{\sec{x}} = x^2\tan{x}\cos{x} = x^2 \frac{\sin{x}}{\cos{x}}\cos{x} = x^2 \sin{x}[/tex]

I'm sure you can take the derivative from there~ :smile:
 
Damn you Data! :smile:
 
got to be quick :-p
 

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