Hi guys, I'm not too sure how to approach this question. Can anyone please give me some tips, etc.?

mathmeme  May 7, 2017

4+0 Answers


\(y\cdot \arctan(e^{3x})=\dfrac{\arccos (x)}{x+2}\) means \(y=\dfrac{\arccos (x)}{(x+2)(\arctan(e^{3x}))}\)

Then use implicit differentiation on the original equation:

\(y\cdot \arctan(e^{3x})=\dfrac{\arccos (x)}{x+2}\\ y'\cdot\arctan(e^{3x})+y\cdot\dfrac{1}{1+e^{6x}}\cdot (3e^{3x})=\dfrac{\left(-\frac{1}{\sqrt{1-x^2}}\right)(x+2)-\arccos x}{(x+2)^2}\)

Left hand side use product rule and chain rule and then right hand side use quotient rule(Wow I used every of them)


Gradient of curve y at point (x_1, y_1)= \(\dfrac{dy}{dx}|_{x=x_1}\)

You substitute \(y=\dfrac{\arccos (x)}{(x+2)(\arctan(e^{3x}))}\) and then express y' in terms of x.

\(y'\cdot\arctan(e^{3x})+y\cdot\dfrac{1}{1+e^{6x}}\cdot (3e^{3x})=\dfrac{\left(-\frac{1}{\sqrt{1-x^2}}\right)(x+2)-\arccos x}{(x+2)^2}\\ y'\cdot\arctan(e^{3x})+\dfrac{\arccos(x)}{(x+2)(\arctan(e^{3x}))}\cdot\dfrac{1}{1+e^{6x}}\cdot (3e^{3x})=\dfrac{\left(-\frac{1}{\sqrt{1-x^2}}\right)(x+2)-\arccos x}{(x+2)^2}\\ y'\cdot\arctan(e^{3x})=\dfrac{\left(-\frac{1}{\sqrt{1-x^2}}\right)(x+2)-\arccos x}{(x+2)^2}-\dfrac{\arccos(x)}{(x+2)(\arctan(e^{3x}))}\cdot\dfrac{1}{1+e^{6x}}\cdot (3e^{3x})\\ y'=\dfrac{\dfrac{\left(-\frac{1}{\sqrt{1-x^2}}\right)(x+2)-\arccos x}{(x+2)^2}-\dfrac{\arccos(x)}{(x+2)(\arctan(e^{3x}))}\cdot\dfrac{1}{1+e^{6x}}\cdot (3e^{3x})}{\arctan(e^{3x})}\)

Then you substitute x = 0 for the last step. Do you want me to do it for you? :D

P.S. : I am an Asian

MaxWong  May 7, 2017

You legend!

mathmeme  May 7, 2017

OK nvm I do it for you XD

\(y'=\dfrac{\dfrac{\left(-\frac{1}{\sqrt{1-x^2}}\right)(x+2)-\arccos x}{(x+2)^2}-\dfrac{\arccos(x)}{(x+2)(\arctan(e^{3x}))}\cdot\dfrac{1}{1+e^{6x}}\cdot (3e^{3x})}{\arctan(e^{3x})}\\ \text{ when x = 0, }y'= \dfrac{\dfrac{\left(-\frac{1}{\sqrt{1-0^2}}\right)(0+2)-\arccos (0)}{(0+2)^2}-\dfrac{\arccos(0)}{(0+2)(\arctan(e^{3(0)}))}\cdot\dfrac{1}{1+e^{6(0)}}\cdot (3e^{3(0)})}{\arctan(e^{3(0)})}\)

The numerator equals -2 - pi/8 and the denominator equals pi/4 <-- that's basic trig and algebra

So the final answer is:

The gradient of the curve is \(-\dfrac{8}{\pi}-\dfrac{1}{2}\) when x = 0

MaxWong  May 7, 2017

Very impressive Max   cool

Melody  May 7, 2017

24 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details