# Using Implicit Differentiation

1. Oct 19, 2006

Alright I have the question:

Find dy/dx by implicit differentiatin

ysin(x^2) = xsin(y^2)

Basically you jus take the derivative of both sides and solve for dy/dx, but I was unsure whether or not my differentation was right. If someone could just get me started in the right direction for this equation I'd be thankful because I'm sure I can finish the rest.

[ysin(x^2)]' = [xsin(y^2)]'
ycos(x^2)*2 = xcos(y^2)*2

Is what I tried but I know that is completely wrong because you have to use the chain rule for both sides. I feel so silly right now because the other examples make more sense, yet this one doesn't haha. Thanks a lot guys.

2. Oct 19, 2006

### BSMSMSTMSPHD

Here's the chain rule. If you are taking the derivative of a function of x with respect to x, then it's just the deivative.

$$\frac{d}{dx} (f(x)) = f'(x)$$

But, if it's a function of y, you have to do the following:

$$\frac{d}{dx} (f(y)) = \frac{d}{dy} (f(y)) \cdot \frac{dy}{dx}$$

In English, this means, "take the derivative with respect to y, and then multiply by dy/dx (which is what you're solving for).

Now, in addition to this rule, you will need to employ the product rule on each side. See if you can try this.

3. Oct 21, 2006

ysin(x^2) = xsin(y^2)
ycos(x^2)*2x+ sin(x^2)*dy/dx = xcos(y^2)*2y*dy/dx+ sin(y^2)*1

Now I'm having trouble simplifying both sides of the equation, and solving for dy/dx. What is the next step in sovling this problem, afterwards it's fairly simple.

4. Oct 21, 2006

Get the $$y'$$ on the same side and factor.

5. Oct 21, 2006

So what you are saying is move the dy/dx to the other side, and then move the other equation to the other side and factor it, so:
ycos(x^2)*2x+ sin(x^2)*dy/dx = xcos(y^2)*2y*dy/dx+ sin(y^2)*1=
ycos(x^2)*2x -sin(y^2) = xcos(y^2)*2y - sin(x^2) dy/dx

And then factor this expression?

6. Oct 21, 2006

$$2xy\cos(x^{2}) + \sin(x^{2})y' - 2xy\cos(y^{2})y' = \sin(y^{2})$$

$$2xy\cos(x^{2}) + y'(sin(x^{2})-2xy\cos(y^{2})) = \sin(y^{2})$$

7. Oct 21, 2006

Ahhhh alright, that makes a whole lot more sense.
Hence: y' = sin(y^2) - 2xycos(x^2) / sin(x^2)-2xycos(y^2)

I suppose writing it using the y' helps a bit more too in simpifying the expression. Alright thanks a lot.

8. Oct 21, 2006

Alright I've got one more question for you if you don't mind, I'm currently working on the question:

If g(x) = secx, find g'''(pi/4)
g'(x) = secxtanx
g''(x) = secx*sec^2x+tanx*secxtanx
= secxtan^2x+sec^3x

Using trig identity: 1+tan^2x = sec^2x
tan^2x=sec^2x-1

Sub in
g'' = sec^3x+secx(sec^2x-1)
=sec^2x+sec^3x-secx
=2sec^3x-secx

I've made it this far, now I need to solve for g'''(x). Do you bring the 3 out in front and solve for:
6*secx-secx using the product rule? Or what do you do from here? Thanks guys. You've been a great help.

9. Oct 21, 2006

$$g''(x) = 2\sec^{3} x - \sec x$$

$$g'''(x) = 3(2)(\sec x)^{2})(\sec x \tan x) - \sec x \tan x$$

$$g'''(x) = 6\sec^{3}x \tan x - \sec x \tan x$$

10. Oct 21, 2006

Ahhh beautiful, thanks a lot, and then when solving for g'''(pi/4), you just sub in (pi/4) for x and get:
6sec^3(pi/4)tan(pi/4)-sec(pi/4)tan(pi/4)
=6(root2)^3*1-(root2)*1
=6(2root2)-(root2)
=12root2-root2
= 11root2 as the final answer I believe.

11. Oct 21, 2006

yes that is correct

12. Oct 21, 2006

Alright thanks a bunch! :D

13. Oct 21, 2006

Alright I just finished another problem and was hoping to see if I just got it right.

The question asked use implicit differentiation to find an equation of the tangent line to the curve at a given point:
x^2+2xy-y^2+x = 2 (1,2)
d/dx(x^2+2xy-y^2+x) = d/dx (2)
2x+2(xy'+(1)y) -2yy' + 1 = 0
2xy'-2yy' = -2x-2y-1
y' [2x-2y] = -2x-2y-1
y' = -2x-2y-1 / 2x-2y
= -2(1)-2(2)-1 / 2(1)-2(2) = 7/2
y-2=7/2(x-1)
y = 7/2x+3/2