# Implicit differentiation

• Nov 14th 2009, 05:55 PM
Archduke01
Implicit differentiation
Finding dy/dx for $\displaystyle y^2 = 5x^3$. Find equation of the tangent line to the graph at the given point. (1, sqrt 5)

Here's what I did;
$\displaystyle 2yy' = 15x^2$
$\displaystyle y' = 15x^2 / 2y$

Where did I go wrong? Or, if it's right by some chance, how do I proceed to find the equation?
• Nov 14th 2009, 06:06 PM
Prove It
Quote:

Originally Posted by Archduke01
Finding dy/dx for $\displaystyle y^2 = 5x^3$

Here's what I did;
$\displaystyle 2yy' = 15x^2$
$\displaystyle y' = 15x^2 / 2y$

Where did I go wrong?

Do you have to use Implicit Differentiation?

$\displaystyle y^2 = 5x^3$

$\displaystyle y = \sqrt{5}x^{\frac{3}{2}}$

$\displaystyle \frac{dy}{dx} = \frac{3\sqrt{5}}{2}x^{\frac{1}{2}}$

$\displaystyle \frac{dy}{dx} = \frac{3\sqrt{5x}}{2}$.

If you WERE going to use Implicit Differentiation:

$\displaystyle y^2 = 5x^3$

$\displaystyle \frac{d}{dx}(y^2) = \frac{d}{dx}(5x^3)$

$\displaystyle \frac{d}{dy}(y^2)\,\frac{dy}{dx} = 15x^2$

$\displaystyle 2y\,\frac{dy}{dx} = 15x^2$

$\displaystyle \frac{dy}{dx} = \frac{15x^2}{2y}$

And since $\displaystyle y = \sqrt{5x^3}$

$\displaystyle \frac{dy}{dx} = \frac{15x^2}{2\sqrt{5x^3}}$

$\displaystyle = \frac{15x^2\sqrt{5x^3}}{10x^3}$

$\displaystyle = \frac{3\sqrt{5x^3}}{2x}$

$\displaystyle = \frac{3x\sqrt{5x}}{2x}$

$\displaystyle = \frac{3\sqrt{5x}}{2}$

which is the same as found above...
• Nov 14th 2009, 06:12 PM
Archduke01
Quote:

Originally Posted by Prove It
And since $\displaystyle y = \sqrt{5x^3}$

Sorry, I just editted my post because I forgot to include that the problem already had given points.
• Nov 14th 2009, 06:20 PM
Prove It
Well if you have $\displaystyle (x,y) = (1, \sqrt{5})$, then what does

$\displaystyle \frac{dy}{dx} = \frac{15x^2}{2y}$ equal?
• Nov 14th 2009, 06:26 PM
Archduke01
Quote:

Originally Posted by Prove It
Well if you have $\displaystyle (x,y) = (1, \sqrt{5})$, then what does

$\displaystyle \frac{dy}{dx} = \frac{15x^2}{2y}$ equal?

$\displaystyle 15/ (2 sqrt5)$ ... which I'm assuming is the slope of the tangent line. How do I get the rest of the equation?
• Nov 14th 2009, 06:29 PM
Prove It
Quote:

Originally Posted by Archduke01
$\displaystyle 15/ (2 sqrt5)$ ... which I'm assuming is the slope of the tangent line. How do I get the rest of the equation?

Correct, which I would rewrite as $\displaystyle \frac{3\sqrt{5}}{2}$ by rationalising the denominator.

Now, you have the tangent line $\displaystyle y = mx + c$.

You know $\displaystyle y = \sqrt{5}, x = 1, m = \frac{3\sqrt{5}}{2}$.

Solve for $\displaystyle c$.
• Nov 14th 2009, 06:33 PM
Archduke01
Quote:

Originally Posted by Prove It
You know $\displaystyle y = 2\sqrt{5}$.

Shouldn't it be $\displaystyle y = sqrt {5}$?
• Nov 14th 2009, 06:34 PM
Prove It
Quote:

Originally Posted by Archduke01
Shouldn't it be $\displaystyle y = sqrt {5}$?

Yes it is, typo.

Will edit now.
• Nov 14th 2009, 06:43 PM
Archduke01
If I did it correctly, the value of c should be $\displaystyle - sqrt 5 / 2$...

So the equation should be $\displaystyle y = 3 sqrt 5 /2 x - sqrt 5/2$

Does that seem right? The answer book says $\displaystyle 15x - 2 (sqrt5)y - 5 = 0$
• Nov 14th 2009, 06:44 PM
Prove It
Quote:

Originally Posted by Archduke01
If I did it correctly, the value of c should be $\displaystyle - sqrt 5 / 2$...

So the equation should be $\displaystyle y = 3 sqrt 5 /2 x - sqrt 5/2$

Does that seem right? The answer book says $\displaystyle 15x - 2 (sqrt5)y - 5 = 0$

Yes it is correct. You just need some algebraic manipulation.
• Nov 14th 2009, 06:46 PM
Archduke01
Quote:

Originally Posted by Prove It
Yes it is correct. You just need some algebraic manipulation.

The equation is correct, or the value of c?
• Nov 14th 2009, 06:46 PM
Prove It
Both.
• Nov 14th 2009, 06:47 PM
Archduke01
Quote:

Originally Posted by Prove It
Both.

Thank you! Sorry for the lengthy back and forth.