find the derivative of the function: x arctan x=e^y at the point (1, ln pi/4)

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find the derivative of the function: x arctan x=e^y at the point (1, ln pi/4)

Calculus1
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to differentiate x*arctan(x) (w.r.t x) you nee to apply product rule to differentiate e^y (w.r.t x) use chain rule
you could make x=e^y ln(x)=y
@sweetburger what do you mean?

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converting exponential to logarthimic they are equivalent... unless I am wrong idk
just trying to help
a^y=x equal loga(x)=y
\[x \arctan(x)=e^y \\ \text{ differentiate both sides w.r.t. } x \\ \frac{d}{dx}(x \arctan(x))= \frac{d}{dx}(e^y) \\ \] oh are you saying she can write: \[\ln(x \arctan(x))=y \text{ instead of } x \arctan(x)=e^y\] ok I was just confused because we don't have the equation x=e^y
well my bad I completely read teh question wrong I thought it was 2 different equations that we were taking the derivative of... my bad
I don't understand what (w.r.t.x) means
with respect to x
ohh ok so find the derivative of x arctanx x and then e^y the apply the product rule and then the chain rule I am looking for the inverse tho
so wrtx is basically differentiating both sides with respect to x or (d/dx) of both sides?
\[\frac{d}{dx} \arctan(x)=\frac{1}{1+x^2}\]
is that what you mean @meaghan25
derivative of that inverse function
\[\frac{d}{dx}e^x=e^x \\ \frac{d}{dx}e^u=\frac{du}{dx} e^u \text{ where} u=u(x)\]
now all you have to really do is replace u with y and apply that product rule I was talking about
yes I need the inverse
yes d/dx means differentiating w.r.t x
what do you need the inverse of ?
sorry I hate open study cus of the typing issue I have with it but the directions say finding dy/dx at a point. in exercises 19 -22, find dy/dx at the given point for the equation 21. x arctan x = e^y , (1, lnpi/4) does that mean I need to find the inverse of the function or just the derivative of dy/dx.
the question says nothing about finding the inverse of anything \[x \arctan(x)=e^y \\ \text{ differentiate both sides w.r.t. x} \\ \frac{d}{dx} x \arctan(x)=\frac{d}{dx}e^y \\ \text{ apply product rule on left hand side } \\ x \frac{d}{dx} \arctan(x)+\arctan(x) \frac{d}{dx} x=\frac{d}{dx}e^y \\ \text {apply chain rule to right hand side } \\ x \frac{d}{dx} \arctan(x)+\arctan(x)\frac{d}{dx}x=\frac{dy}{dx} e^y\]
dy/dx means find the derivative of the y w.r.t. x
you need to replace d/dx arctan(x) and replace d/dx x with arctan(x)'s derivative and x's derivative respectively
anyways once you are done enter in your point and solve for dy/dx and you are done
ok thank you so much
do you want me to check your answer?
or even your work
no I have the answer already just couldn't figure out what I needed to do exactly
alright
Try this online calculator to solve the problem http://www.acalculator.com/quadratic-equation-calculator-formula-solver.html I hope it is helpful.

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