Exp/Log Differentiation

[reardear]

A cruve is described by \(\displaystyle y^{2} = x^{log_2(x)}\) with domain \(\displaystyle D = \{ {x} \in \mathbb{R} \backslash x > 0 \}\)

a) Determine the y-values, for which x=4.
I got y = 4 and -4

b) Determine equation(s) for the tangent to the given curve at x=4.
I think I do implicit differentiation here and then plug into point-slope equation? If so, this is what I got so far.. I'm not very good with these. Some help on how to differentiate here would be appreciated!

\(\displaystyle lny^{2} = lnx^{log_2(x)}\)

\(\displaystyle 2lny = lnx^\frac{\mathrm{1} }{\mathrm{xln} 2}\)

\(\displaystyle 2\frac{1}{y} = \frac{\mathrm{1} }{\mathrm{xln} 2}lnx\)

c) The normal is defined as the line perpendicular to a tangent at a specified point on a curve. Determine the equation(s) for the normal to the given curve at x=4.
Do I find the perpendicular equation of b?

 

[daon2]

A cruve is described by \(\displaystyle y^{2} = x^{log_2(x)}\) with domain \(\displaystyle D = \{ {x} \in \mathbb{R} \backslash x > 0 \}\)

a) Determine the y-values, for which x=4.
I got y = 4 and -4

b) Determine equation(s) for the tangent to the given curve at x=4.
I think I do implicit differentiation here and then plug into point-slope equation? If so, this is what I got so far.. I'm not very good with these. Some help on how to differentiate here would be appreciated!

\(\displaystyle lny^{2} = lnx^{log_2(x)}\)

\(\displaystyle 2lny = lnx^\frac{\mathrm{1} }{\mathrm{xln} 2}\)

\(\displaystyle 2\frac{1}{y} = \frac{\mathrm{1} }{\mathrm{xln} 2}lnx\)

c) The normal is defined as the line perpendicular to a tangent at a specified point on a curve. Determine the equation(s) for the normal to the given curve at x=4.
Do I find the perpendicular equation of b?

I agree with your choice to use logarithmic differentiation. However, it looks like you tried to take the derivative of the power of x for some reason. Maybe this will help:

\(\displaystyle \ln x^{\log_2(x)} = \log_2(x)\ln(x) = \dfrac{\left(\ln(x)\right)^2}{\ln(2)}\)

Note this is before any differentiation is done.

Also, it looks like you are having trouble with the chain rule, when you take the derivative of \(\displaystyle \ln y\). You are assuming y depends on x, so that for any function \(\displaystyle f(y)\) (in your case f(y)=2ln(y)), you have \(\displaystyle \dfrac{d}{dx}f(y) = \left(\dfrac{d}{dy}f(y)\right)\cdot \underbrace{\dfrac{d}{dx}y}_{y'}\)

 

[reardear]

I totally forgot about that rule! Thanks. I also received some extra help and I think I figured it out now!