# Thread: Implicit differentiation and tangent lines

1. ## Implicit differentiation and tangent lines

One of these I know is wrong ,one of these I don't understand how to do.

1) Use logarithmic differentiation to find the derivative of the function.
ln(y) = ln(x)ln(sin(x))
F(x) = ln(x)ln(sin(x))
f(x) = ln(x); f'(x) = 1/x
g(x) = ln(sin(x))
g'(x) = cosx/sinx

y'/y = [ln(sin(x))/x] + [ln(x)cos(x)/sin(x)]
y' = y{[ln(sin(x))/x] + [ln(x)cos(x)/sin(x)]}

y' = [sin(x)^ln(x)ln(sin(x))/x] + (ln(x)^2) * cos(x)

Not sure where i went wrong on that one.

2) The curve with equation y2 = x3 + 3x2 is called the Tschirnhausen cubic. -Find an equation of the tangent line to this curve at the point (1, -2).
-Find the points on the curve where the tangent line has a vertical asymptote

I was under the impression I had to derive the function, and then find points where it is undefined, but the question is asking for y, not y'.

2. Originally Posted by CFem
One of these I know is wrong ,one of these I don't understand how to do.

1) Use logarithmic differentiation to find the derivative of the function.
ln(y) = ln(x)ln(sin(x))
F(x) = ln(x)ln(sin(x))
f(x) = ln(x); f'(x) = 1/x
g(x) = ln(sin(x))
g'(x) = cosx/sinx

y'/y = [ln(sin(x))/x] + [ln(x)cos(x)/sin(x)]
y' = y{[ln(sin(x))/x] + [ln(x)cos(x)/sin(x)]}

y' = [sin(x)^ln(x)ln(sin(x))/x] + (ln(x)^2) * cos(x)

Not sure where i went wrong on that one.

2) The curve with equation y2 = x3 + 3x2 is called the Tschirnhausen cubic. -Find an equation of the tangent line to this curve at the point (1, -2).
-Find the points on the curve where the tangent line has a vertical asymptote ... recheck this statment

I was under the impression I had to derive the function, and then find points where it is undefined, but the question is asking for y, not y'.
$\displaystyle y = (\sin{x})^{\ln{x}}$

$\displaystyle \ln{y} = \ln\left[(\sin{x})^{\ln{x}}\right]$

$\displaystyle \ln{y} = \ln{x} \cdot \ln(\sin{x})$

$\displaystyle \frac{y'}{y} = \ln{x} \cdot \cot{x} + \ln(\sin{x}) \cdot \frac{1}{x}$

$\displaystyle y' = (\sin{x})^{\ln{x}}\left[\ln{x} \cdot \cot{x} + \ln(\sin{x}) \cdot \frac{1}{x}\right]$

$\displaystyle y^2 = x^3 + 3x^2$

$\displaystyle 2y \cdot y' = 3x^2 + 6x$

$\displaystyle y' = \frac{3x(x+2)}{2y}$

at $\displaystyle (1,-2)$ , $\displaystyle m = -\frac{9}{4}$

$\displaystyle y + 2 = -\frac{9}{4}(x - 1)$

tangent lines are vertical when $\displaystyle y'$ is undefined ...

$\displaystyle y' = \frac{3x(x+2)}{2y}$ is undefined when $\displaystyle y = 0$

points on the curve where $\displaystyle y = 0$ ...

$\displaystyle 0 = x^3 + 3x^2$

$\displaystyle 0 = x^2(x + 3)$

$\displaystyle x = 0$ , $\displaystyle x = -3$

points where the tangent line is vertical are (0,0) and (-3,0)

3. #2 is wrong.

Thanks for the help so far, though.

Edit: And I was wrong, the question should read "The curve with equation y2 = x3 + 3x2 is called the Tschirnhausen cubic. Find an equation of the tangent line to this curve at the point (1, -2)

At what points does this curve have a horizontal tangent?"

However both parts were still maked incorrect, and the points are supposed to have different y values.

4. Originally Posted by CFem

Edit: And I was wrong, the question should read "The curve with equation y^2 = x^3 + 3x^2 is called the Tschirnhausen cubic. Find an equation of the tangent line to this curve at the point (1, -2)

At what points does this curve have a horizontal tangent?"
first, note how to denote exponents in your posts by looking at the correction above.

the derivative is the same as completed above.

substitute the given coordinates (1,-2) into the derivative to determine the slope of the tangent line ... then use the point-slope form of a linear equation to find the tangent line equation.

a horizontal tangent line occurs where $\displaystyle \frac{dy}{dx}$ = 0

set the derivative equal to 0, solve for x in terms of y or y in terms of x (whichever is easier), and substitute the result into the original equation for the curve ... solve.

5. Sorry about the notation. Was copy and pasting.

y = (-9/4x) -1/4

Solving dy/dx for x gave me -2 (marked correct) and 0 instead of -3 and 0.

Which gave me the points:

(0,-1/4)
(-2,17/4)

Am I mistaken? Still is getting refused (except for the -2)

Edit: Nevermind, I lost a sign. I got (-9/4)x + (1/4)

But the only coordinate that isn't being marked off is -2. -2 in that equation gives 19/4, correct? It's also still refusing to accept 0 as a valid point.

Edi2: I don't think I'm using the right equation. It doesn't look like subbing the x values into y' gives the proper number. Should the number be substituted into the original equation?