1. ## Implicit differentiation

Finding dy/dx for $y^2 = 5x^3$. Find equation of the tangent line to the graph at the given point. (1, sqrt 5)

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

Where did I go wrong? Or, if it's right by some chance, how do I proceed to find the equation?

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

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

Where did I go wrong?
Do you have to use Implicit Differentiation?

$y^2 = 5x^3$

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

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

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

If you WERE going to use Implicit Differentiation:

$y^2 = 5x^3$

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

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

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

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

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

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

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

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

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

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

which is the same as found above...

3. Originally Posted by Prove It
And since $y = \sqrt{5x^3}$
Sorry, I just editted my post because I forgot to include that the problem already had given points.

4. Well if you have $(x,y) = (1, \sqrt{5})$, then what does

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

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

$\frac{dy}{dx} = \frac{15x^2}{2y}$ equal?
$15/ (2 sqrt5)$ ... which I'm assuming is the slope of the tangent line. How do I get the rest of the equation?

6. Originally Posted by Archduke01
$15/ (2 sqrt5)$ ... 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 $\frac{3\sqrt{5}}{2}$ by rationalising the denominator.

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

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

Solve for $c$.

7. Originally Posted by Prove It
You know $y = 2\sqrt{5}$.
Shouldn't it be $y = sqrt {5}$?

8. Originally Posted by Archduke01
Shouldn't it be $y = sqrt {5}$?
Yes it is, typo.

Will edit now.

9. If I did it correctly, the value of c should be $- sqrt 5 / 2$...

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

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

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

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

Does that seem right? The answer book says $15x - 2 (sqrt5)y - 5 = 0$
Yes it is correct. You just need some algebraic manipulation.

11. Originally Posted by Prove It
Yes it is correct. You just need some algebraic manipulation.
The equation is correct, or the value of c?

12. Both.

13. Originally Posted by Prove It
Both.
Thank you! Sorry for the lengthy back and forth.