How do you differentiate $\mathrm{tan}2x=\mathrm{cos}3y$

amacorrit80
2022-07-18
Answered

How do you differentiate $\mathrm{tan}2x=\mathrm{cos}3y$

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asked 2022-08-14

Use implicit differentiation to calculate $\frac{dy}{dx}$ for ${x}^{2}+xy-6={y}^{3}$

asked 2022-07-14

I am trying to implicitly differentiate

$\mathrm{sin}(x/y)=1/2$

The solution manual says

Step 1.

$\mathrm{cos}(x/y)\cdot \frac{y-x\frac{dy}{dx}}{{y}^{2}}=0$

But I don't understand how they arrive at this next part:

Step 2.

$y-x\frac{dy}{dx}=0$

Is $\mathrm{cos}(x/y)={y}^{2}$?

$\mathrm{sin}(x/y)=1/2$

The solution manual says

Step 1.

$\mathrm{cos}(x/y)\cdot \frac{y-x\frac{dy}{dx}}{{y}^{2}}=0$

But I don't understand how they arrive at this next part:

Step 2.

$y-x\frac{dy}{dx}=0$

Is $\mathrm{cos}(x/y)={y}^{2}$?

asked 2022-07-21

The equation $2{x}^{4}{z}^{2}+{y}^{4}={x}^{2}-4x{y}^{5}z$ defines $z$ implicitly as a function of $x$ and $y$. Find $\frac{\mathrm{\partial}z}{\mathrm{\partial}x}$ at the point $(1,-1,0)$

The solution was as follows:

So if we derive with respect to $x$ we get

$8{x}^{3}{z}^{2}+4{x}^{4}z\frac{\mathrm{\partial}z}{\mathrm{\partial}x}=2x-4{y}^{5}z-4x{y}^{5}\frac{\mathrm{\partial}z}{\mathrm{\partial}x}$

Plugging inn $(1,-1,0)$ we get $\frac{\mathrm{\partial}z}{\mathrm{\partial}x}=\frac{1}{2}$.

But when we differentiate with respect to $x$, on the last expression they used the chain rule on $x$ and $z$ and treated $y$ as a constant. Shouldn't you also use the chain rule on $y$ and get something like $-4x{y}^{5}z-4x{y}^{5}\frac{\mathrm{\partial}z}{\mathrm{\partial}x}-20x{y}^{4}z\frac{\mathrm{\partial}y}{\mathrm{\partial}x}$.

I realize now that since $z=0$ in our point, the last expression would actually fall away and we would get the right answer. But the solution doesn't contain the last expression at all, so I'm confused about whether or not I've misunderstood implicit differentiation.

The solution was as follows:

So if we derive with respect to $x$ we get

$8{x}^{3}{z}^{2}+4{x}^{4}z\frac{\mathrm{\partial}z}{\mathrm{\partial}x}=2x-4{y}^{5}z-4x{y}^{5}\frac{\mathrm{\partial}z}{\mathrm{\partial}x}$

Plugging inn $(1,-1,0)$ we get $\frac{\mathrm{\partial}z}{\mathrm{\partial}x}=\frac{1}{2}$.

But when we differentiate with respect to $x$, on the last expression they used the chain rule on $x$ and $z$ and treated $y$ as a constant. Shouldn't you also use the chain rule on $y$ and get something like $-4x{y}^{5}z-4x{y}^{5}\frac{\mathrm{\partial}z}{\mathrm{\partial}x}-20x{y}^{4}z\frac{\mathrm{\partial}y}{\mathrm{\partial}x}$.

I realize now that since $z=0$ in our point, the last expression would actually fall away and we would get the right answer. But the solution doesn't contain the last expression at all, so I'm confused about whether or not I've misunderstood implicit differentiation.

asked 2022-08-18

How does implicit differentiation work?

asked 2022-08-14

What is the derivative of $y=6xy$

asked 2022-07-22

I've been stuck on a certain implicit differentiation problem that I've tried several times now.

$\frac{{x}^{2}}{x+y}={y}^{2}+6$

I know to take the derivatives of both sides and got:

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

I blackuced that to get:

$2{x}^{2}+2xy-{x}^{2}-{x}^{2}\ast dy/dx=(2y\ast dy/dx)(x+y{)}^{2}$

I then divided both sides by (2y*dy/dx) and multiplied each side by the reciprocals of the first three terms of the left side. Then I factoblack dy/dx out of the left side and multiplied by the reciprocal of what was left to get dy/dx by itself. I ended up with:

$dy/dx=(2y(x+y{)}^{2})/(4{x}^{7}y)$

but this answer was wrong. I only have one more attempt on my online homework and I can't figure out where I went wrong. Please help!

$\frac{{x}^{2}}{x+y}={y}^{2}+6$

I know to take the derivatives of both sides and got:

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

I blackuced that to get:

$2{x}^{2}+2xy-{x}^{2}-{x}^{2}\ast dy/dx=(2y\ast dy/dx)(x+y{)}^{2}$

I then divided both sides by (2y*dy/dx) and multiplied each side by the reciprocals of the first three terms of the left side. Then I factoblack dy/dx out of the left side and multiplied by the reciprocal of what was left to get dy/dx by itself. I ended up with:

$dy/dx=(2y(x+y{)}^{2})/(4{x}^{7}y)$

but this answer was wrong. I only have one more attempt on my online homework and I can't figure out where I went wrong. Please help!

asked 2022-07-23

I was looking to implicitly differentiate

$-22{x}^{6}+4{x}^{33}y+{y}^{7}=-17$

and found it to be

$\frac{dy}{dx}}={\displaystyle \frac{132{x}^{5}-132{x}^{32}y}{4{x}^{33}+7{y}^{6}}$

Now, I am trying to find the equation of the tangent line to the curve at the coordinate (1,1). So I then plug both 1 in for x and y into the above equation and come up with

$\frac{0}{11}$

Now I go to solve

$y-y1=m(x-x1)$

getting

$y-1=0(x-1)$

resulting in $y=1$ and the equation to be $y=x+1$ for my final answer. Am I going about this in the correct manner?

$-22{x}^{6}+4{x}^{33}y+{y}^{7}=-17$

and found it to be

$\frac{dy}{dx}}={\displaystyle \frac{132{x}^{5}-132{x}^{32}y}{4{x}^{33}+7{y}^{6}}$

Now, I am trying to find the equation of the tangent line to the curve at the coordinate (1,1). So I then plug both 1 in for x and y into the above equation and come up with

$\frac{0}{11}$

Now I go to solve

$y-y1=m(x-x1)$

getting

$y-1=0(x-1)$

resulting in $y=1$ and the equation to be $y=x+1$ for my final answer. Am I going about this in the correct manner?