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A particle is moving along the curve $y=\dfrac{e^x-e^{-x}}{e^x+e^{-x}}$. As the particle passes through the point $x=\ln2$, its x-coordinate increases at a rate of 5cm/sec. How fast is the slope of the tangent line to the curve $y=\dfrac{e^x-e^{-x}}{e^x+e^{-x}}$ at the particle changing at that instant?

This is an exercise question of my brother, but I did not quite understand the English. Am I supposed to find $d^2y/dx^2$ at that point or just calculate the first derivative? What does "how fast does the slope of the graph of the function change?" mean? Should I bring in a new time variable $t$, also?

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  • $\begingroup$ If you want the rate at which the slope changes, you need the second derivative. $\endgroup$
    – saulspatz
    Aug 15, 2020 at 16:09
  • $\begingroup$ The variable $x$ is a function of $t$, so you need to apply the Chain Rule. $\endgroup$
    – N. F. Taussig
    Aug 15, 2020 at 16:10
  • $\begingroup$ "Should I bring in a new time variable $t$, also?" — Yes. The clue for this is in the sentence that says $x$ is increasing at $5$ cm/sec at $x = \ln 2$. $\endgroup$
    – Théophile
    Aug 15, 2020 at 16:12
  • $\begingroup$ "How fast blah is changing" means "derivative of blah". "The slope of the tangent line" means "the first derivative of the function". So "how fast the slope of the tangent line is changing" means "the derivative of the slope of the tangent line" mean "the derivative of the first derivative" means "the second derivative". $\endgroup$
    – fleablood
    Aug 15, 2020 at 16:32

1 Answer 1

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Find the slope in terms of x and differentiate it with respect to t ,

As you know dx/dt you will get rate of change of slope

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  • $\begingroup$ In flow I did it mistakenly , thanks for it $\endgroup$
    – Namburu Karthik
    Aug 15, 2020 at 16:16

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