Understanding Implicit Differentiation: Solving for Second Order Derivatives

Duncan1382
Messages
9
Reaction score
0
Find d^2/dx^2(3y^2+8y=3x)

I managed to get dy/dx = 3 / (6y + 8) but I have no clue where to go from here.

According to WolfRamAlpha, the answer is -27/(4(16 + 9x)(4 + 3y)), but since dy/dx doesn't have any x value in it, I don't see how the derivative of it would.

I've played around with it for a long time, and I just can't get it. Help please?
 
Physics news on Phys.org
Do you need to express in terms of x and y? I don't see what's wrong if you express it in y which is what i would do
 
semc said:
Do you need to express in terms of x and y? I don't see what's wrong if you express it in y which is what i would do

You can't express y in terms of x. It's not a function. You can express it in terms of x and y, but that just makes the math longer.
 
Duncan - WolfRamAlpha mad a substitution in the denominator using the original expression.

Firstly - did you compute the second derivative correctly? You should have yielded:

<br /> y&#039; =\frac{3}{8+6 \cdot y}<br />
and
<br /> y&#039;&#039; = -\frac{6 \cdot (y&#039;)^2}{8+6 \cdot y}<br />

Now, using the first into the second

<br /> y&#039;&#039; = -\left (\frac{3}{8+6 \cdot y} \right)^2 \left (\frac{6}{8+6 \cdot y} \right ) = \frac{-54}{8 \cdot (4+3 \cdot y)^3}<br />

Now, using the fact that 3y^2+8y=3x, you need to show yourself that (4+3y)^2 = 9x +16. Use this in above to yield result.
 
Oh. Now I get it.

Thank you.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top