Finding the equation to a tangent line of ln(xy)
I am having an issue solving this problem on my final prep sheet. It seems I need to use implicit differentiation because of the two variables but I am not exactly sure if that is correct.
"Find the equation of the tangent line to ln(xy) = 2x at the point (1, e^2)."
I assumed first you subtract the 2x to the left side and then implicitly differentiate but I am getting completely lost in it.
$\endgroup$2 Answers
$\begingroup$An idea: assuming the implict equation is
$$2x=\log xy=\log x+\log y\implies \log y=2x-\log x\implies y=\frac{e^{2x}}x$$
and now you can differentiate:
$$y'=\frac{2xe^{2x}-e^{2x}}{x^2}\;\;\text{and etc.}$$
$\endgroup$ $\begingroup$As $\ln(xy)=\ln x+\ln y$ for $x,y>0$
we write $$2x=\ln x+\ln y$$
We know $\displaystyle\frac{dy}{dx}$ represents the gradient of the tangent
So, differentiating wrt $x$ $$2=\frac1x+\frac1y\frac{dy}{dx}$$
At $\displaystyle(1,e^2),2=1+\frac1{e^2}\frac{dy}{dx}\implies \frac{dy}{dx}_{\text{ at}(1,e^2)}=e^2$
Do you know how to find the equation of a straight line passing through a given point and having a given gradient
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