# Thread: Implicit differentiation with trig

1. ## Implicit differentiation with trig

1. sin(x+y) = x-y The answer is (1-cos(x+y))/((cos x+y)+1)
derivitive: cos(x+y)(1+dy/dx)=1-dy/dx
(dy/dx)+cos(x+y)(1+dy/dx)=1
(dy/dx)+cos(x+y)+cos(x+y(dy/dx))=1
(dy/dx)+cos(x+y(dy/dx))=1-cos(x+y)
I believe this is when i went wrong or something.

2. cos(xy)=1-x^2 the answer is (2x-y sin (xy))/(x sin (xy))
derivitive: -sin(xy)(x(dy/dx)+y)=-2x
2x-sin(xy)(x(dy/dx)+y)=0
And I got lost from here.

3.ln(xy)=e^2x The answer is (2e^2x-x^-1)y
derivitive: (1/xy)(xy'+y)=2e^2x
(-xy'+x-xy'+y)/(xy)^2=2e^2x
And then this step was when I got lost as to how this translates to the answer.

2. Don't forget the chain rule with these.

#1: $sin(x+y)-x+y=0$

$cos(x+y)(1+\frac{dy}{dx})-1+\frac{dy}{dx}=0$

$cos(x+y)+\frac{dy}{dx}cos(x+y)-1+\frac{dy}{dx}=0$

$\frac{dy}{dx}=\frac{1-cos(x+y)}{cos(x+y)+1}$

#2: $cos(xy)-x^{2}=0$

$(x\frac{dy}{dx}+y)(-sin(xy))+2x=0$

$-xsin(xy)\frac{dy}{dx}-ysin(xy)+2x=0$

$\frac{dy}{dx}=\frac{2x-ysin(xy)}{xsin(xy)}$

The last one.

$ln(xy)-e^{2x}=0$

$\frac{x\frac{dx}{dy}+y}{xy}=2e^{2x}$

$\frac{dy}{dx}=\frac{(2xe^{2x}-1)y}{x}$

The best way to do these is the way earboth showed.

Take the partials with respect to x and y:

$\frac{d}{dx}[ln(xy)-e^{2x}]=\frac{1}{x}-2e^{2x}$

$\frac{d}{dy}[ln(xy)-e^{2x}]=\frac{1}{y}$

$-\frac{\frac{1}{x}-2e^{2x}}{\frac{1}{y}}=\frac{(2xe^{2x}-1)y}{x}$

Same as above. See?.

3. I get 1 and 2. But for 3, the problem i have is how
$\frac{x\frac{dx}{dy}+y}{xy}=2e^{2x}$

Becomes this:

$
\frac{dy}{dx}=\frac{(2xe^{2x}-1)y}{x}
$

Other than that, I get the rest of the material.

4. Originally Posted by driver327
I get 1 and 2. But for 3, the problem i have is how
$\frac{x\frac{dx}{dy}+y}{xy}=2e^{2x}$

Becomes this:

$
\frac{dy}{dx}=\frac{(2xe^{2x}-1)y}{x}
$

Other than that, I get the rest of the material.
Hello,

I'll show you the steps of transformation only:

$\frac{x\frac{dx}{dy}+y}{xy}=2e^{2x}$ . Your equation. Multiply by $(xy)$

$x\frac{dx}{dy}+y=2e^{2x}\cdot xy$ . Subtract y

$x\frac{dx}{dy}=2e^{2x}\cdot xy - y$ . Divide by x

$\frac{dx}{dy}=\frac{2e^{2x}\cdot xy - y}{x}$ . Factor the numerator:

$\frac{dx}{dy}=\frac{\left(2e^{2x}\cdot x - 1\right) y}{x}$