Inverse trigonometric function

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SUMMARY

This discussion focuses on finding derivatives of various functions, particularly those involving inverse trigonometric functions and logarithmic functions. The user requests assistance with specific derivatives, including y = ln((x+1)/(x+2)^3) and y = arctan(x/(a - √(a² - x²))). A key method highlighted is using the relationship between inverse functions and their derivatives, exemplified by differentiating y = ln(x) through the exponential function. Participants emphasize the importance of understanding each step in the differentiation process.

PREREQUISITES
  • Understanding of derivatives and differentiation rules
  • Familiarity with inverse functions, particularly logarithmic and trigonometric
  • Knowledge of the chain rule in calculus
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the differentiation of inverse functions, specifically focusing on y = ln(x) and its relationship with e^x
  • Practice finding derivatives of functions involving arctan and arcsin
  • Explore the application of the chain rule in more complex functions
  • Review examples of derivatives involving logarithmic and trigonometric identities
USEFUL FOR

Students and educators in calculus, mathematicians focusing on differentiation techniques, and anyone seeking to improve their understanding of inverse functions and their derivatives.

noypihenyo
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Find dy
dx


1.) y = In __x2 (x+1)___
(x + 2)3

2.) y = x3 ( 3lnx-1)

3.) y = __cos6 2x__
(1-sin2x)3


4.) y = __tan 2x__
1- cot 2x


5.) y = x e (exponent pa po ng e) sin2x


6.) Arc tan ____x_______
a- √ a2-x2


7.) __x_______ _ Arc sin _x_
a-√ a2-x2 a

_________
8.) Arc tan /__3x - 4_
√ 4
_______
9.) Arc cos ( 1 - _x_+ √ 2ax-x2
a

pls help i need the answer October 8
 
Last edited:
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3) and 4) are not about inverse functions. Show us what you have tried that didn't work.

For the rest, there is a clever trick you can apply for all of them. Say I want to find the derivative of y(x) = ln(x). since ln(x) is the inverse function of [itex]e^x[/itex]=exp(x), I begin by "taking both sides into the argument of exp(x)", such that the equation becomes [itex]e^y=e^{ln(x)}=x[/itex]. And now I differentiate both side wrt x: [itex]\frac{d}{dx}e^y=\frac{d}{dx}x[/itex]. I use the chain rule on the left hand side: [itex]e^y\frac{dy}{dx} = 1[/itex], or, now replacing y by ln(x): [itex]\frac{dy}{dx} = 1/e^{ln(x)}=1/x[/itex]. I suggest you study this exemple until you understand each step. Then, try one of your problems. If you get stuck, you can ask us for help by writing exactly what you've done and where you're stuck.
 

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