Derivative of a surd containing exponential sum

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Homework Help Overview

The discussion revolves around finding the derivative of the function y = sqrt(exp(x) - exp(-x)), which involves exponential functions and their derivatives. Participants are exploring the implications of using hyperbolic functions in relation to the original expression.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants have attempted to differentiate the function using the chain rule and have also explored logarithmic differentiation. There are questions regarding the discrepancies between their results and those provided by a derivative calculator.

Discussion Status

Some participants have noted that their results align with hyperbolic function representations, while others are questioning the validity of the derivative calculator's output. There is ongoing exploration of potential syntax errors or misunderstandings in the differentiation process.

Contextual Notes

Participants are considering the implications of substituting hyperbolic functions for the exponential terms and are discussing the equivalence of different forms of the derivative. There is a focus on ensuring that the approaches taken are consistent with the definitions and properties of the functions involved.

basher87
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Homework Statement


y = sqrt(exp(x) - exp(-x))


Homework Equations


dy/dx = dy/du.du/dx - chain rule
d/dx(exp(x)) = exp(x) - derivative of exp(x)


The Attempt at a Solution



y = sqrt(exp(x) - exp(-x))
y' = (1/2).(1/sqrt(exp(x) - exp(-x))).[exp(x) - (exp(-x).-1)]

y' = [exp(x) + exp(-x)]/[2*sqrt(exp(x) - exp(-x))]

i've tried logarithmic differentation as well coming up with the same answer. derivative calculator says different. please help
 
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basher87 said:

Homework Statement


y = sqrt(exp(x) - exp(-x))

Homework Equations


dy/dx = dy/du.du/dx - chain rule
d/dx(exp(x)) = exp(x) - derivative of exp(x)

The Attempt at a Solution



y = sqrt(exp(x) - exp(-x))
y' = (1/2).(1/sqrt(exp(x) - exp(-x))).[exp(x) - (exp(-x).-1)]

y' = [exp(x) + exp(-x)]/[2*sqrt(exp(x) - exp(-x))]

I've tried logarithmic differentiation as well coming up with the same answer. derivative calculator says different. please help
Hello basher87. Welcome to PF !

What does the derivative calculator say ? Something with cosh and tanh functions ?
 
thankyou sammy

no it comes up with this.

i have checked the hyperbolic functions, in fact i just substituted 2sinhx for exp(x) - exp(-x)

and got the same answer

(exp(-x)*(exp(2.5x) + exp(.5x)))/2(sqrt(exp(2x) - 1) is what the calculator gives
 
basher87 said:
thankyou sammy

no it comes up with this.

i have checked the hyperbolic functions, in fact i just substituted 2sinhx for exp(x) - exp(-x)

and got the same answer

(exp(-x)*(exp(2.5x) + exp(.5x)))/(2(sqrt(exp(2x) - 1)) is what the calculator gives
Is that possibly equivalent to your answer?
 
i solved the equation in terms of the hyperbolic function.

y = sqrt(2sinh x), the derivative calculator gave te same output.

if i solve it in terms of the exponential equation i get the equivalent but the calculator doesnt. Is it possible that it is a syntax error
 
basher87 said:
thankyou sammy

no it comes up with this.

i have checked the hyperbolic functions, in fact i just substituted 2sinhx for exp(x) - exp(-x)

and got the same answer

(exp(-x)*(exp(2.5x) + exp(0.5x)))/2(sqrt(exp(2x) - 1) is what the calculator gives
[itex]\displaystyle \frac{e^{-x}(e^{2.5x} + e^{0.5x})}{2\sqrt{e^{2x} - 1}}[/itex]

[itex]\displaystyle =\frac{e^{-x}e^{1.5x}(e^{x} + e^{-x})}{2\sqrt{e^{2x} - 1}}[/itex]

[itex]\displaystyle =\frac{e^{x} + e^{-x}}{2e^{-0.5}\sqrt{e^{2x} - 1}}[/itex]

[itex]\displaystyle =\frac{e^{x} + e^{-x}}{2\sqrt{e^{x} - e^{-x}}}[/itex]
 

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