Hi guys, need some help on this calculus question :)

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

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

Very impressive Max