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of 4 cos x sin y =1
here's what i did
d/dx 4 cosx sin y)= 1 d/dx
4sinx cosy=0
y'= 4 sinx cosy
but the answer is tan x tan y
here's what i did
d/dx 4 cosx sin y)= 1 d/dx
4sinx cosy=0
y'= 4 sinx cosy
but the answer is tan x tan y
here's what i did 4 d/dx (Cos x) + cos x * d/dx (4) * sin y dy/ dx
is this right or am i wrong with the product rule of this
then i got 4 sin X + sin x * cos y dy/dx
I think that you are misunderstanding the product rule. Firstly, since 4 is a constant, you can take it out of the situation. So
[tex]\frac{d}{dx}4cosxsiny[/tex] = [/tex] 4 \frac{d}{dx}cosxsiny[/tex]
Now, the product rule is [tex] \frac{d}{dx}[f(x)g(x)]=f(x)g'(x) + g(x)f'(x) [/tex] I can demonstrate the proof if you wish but I won't clutter up the post unnecesarily.
I am still editing this, but I see that you are still posting.. PLEASE use the edit button... I am still editing this post so check back in a few minutes...
Okay, anyways... So I will do an example problem and show you how to do it.
Let me explain implicit differentiation for you. When you are differentiating terms involving x alone, you can differentiate as usual. However, when you differentiate terms involving y, you must apply the Chain Rule, because you are assuming that y is defined implicitly as a differentiable function of x.
Lets say the problem was:
find [tex] \frac{dy}{dx}[/tex] given that [tex]y^{3}+y^{2}-5^{y}-x^{2}=-4[/tex]
Solution:
1.Differentiate both sides of the equation with respect to x.
[tex] \frac{d}{dx}[y^{3}+y^{2}-5y-x^{2}]=\frac{d}{dx}[-4] [/tex]
[tex] \frac{d}{dx}[y^{3}] + \frac{d}{dx}[y^{2}] - \frac{d}{dx}[5y] - \frac{d}{dx}[x^{2}] = \frac{d}{dx}[-4][/tex]
[tex]3y^{2}\frac{dy}{dx}+2y\frac{dy}{dx}-5\frac{dy}{dx}-2x=0[/tex] (chain rule)
2. Collect the dy/dx terms on the left side of the equation.
[tex]3y^{2}\frac{dy}{dx}+2y\frac{dy}{dx}-5\frac{dy}{dx}=2x[/tex]
3.Factor dy/dx out of the left side of the equation
[tex]\frac{dy}{dx}(3y^{2}+2y-5)=2x[/tex]
4.Solve for dy/dx by dividing by [tex]3y^{2}+2y-5[/tex]
[tex]\frac{dy}{dx}=\frac{2x}{3y^{2}+2y-5}[/tex]
There you go... Do you understand now?
then i did this - cos x/ sin y - sin x/ cos y = tan x tan y
d/dx (4 cos x * sin y)= 1
d/dx 4 * d/dx cos X * d/dx sin y = d/dx (1)
d/dx 0 * d/dx sin x * d/dx cos y =0
then d/dx( sinx * cosy= 0
Lol, what are you going on about? I am so lost, is this a joke?
is "NO!")afcwestwarrior said:isnt the derivative of cos sin and sin cos