Using Implicit Differentiation

[loadsy]

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.

 

[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.

 

[loadsy]

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.

 

[courtrigrad]

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

 

[loadsy]

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?

 

[courtrigrad]

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

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

 

[loadsy]

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.

 

[loadsy]

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.

 

[courtrigrad]

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

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

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

 

[loadsy]

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.

 

[courtrigrad]

yes that is correct

 

[loadsy]

Alright thanks a bunch! :D

 

[loadsy]

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