Implicit Differentiation Explained: In-Depth Examples, Equations, and Expert Insights

Given $y=y(x)$ what is
$\frac{d}{dx}(e^x(x^2+y^2))$
I think its the $\frac{d}{dx}$ confusing me, I don't what effect it has compablack to $\frac{dy}{dx}$. Any help will be greatly appreciated.

$x\sin <!--\; \; -->(4x+5y)=y\cos <!--\; \; -->(x)$
I am trying to use implicit differentiation to find dx/dy for this problem but the answer i keep getting is
$4x\cos <!--\; \; -->(4x+5y)=-<!--\; -\; -->y\sin <!--\; \; -->(x)$
and I am stuck.

I was wondering whether anybody could explain how you derive this implicit differentiation rule:
$\frac{\mathrm{\partial <!--\; \partial \; -->}z}{\mathrm{\partial <!--\; \partial \; -->}x}=\frac{-<!--\; -\; -->\mathrm{\partial <!--\; \partial \; -->}f/\mathrm{\partial <!--\; \partial \; -->}x}{\mathrm{\partial <!--\; \partial \; -->}f/\mathrm{\partial <!--\; \partial \; -->}z}$
if you have a function $z$ implicity defined by $f(x,y,z)=0$?

The equation $2x^4{z}^2+y^4=x^2-<!--\; -\; -->4x{y}^{5}z$ defines $z$ implicitly as a function of $x$ and $y$. Find $\frac{\mathrm{\partial <!--\; \partial \; -->}z}{\mathrm{\partial <!--\; \partial \; -->}x}$ at the point $(1,-<!--\; -\; -->1,0)$
The solution was as follows:
So if we derive with respect to $x$ we get
$8x^3{z}^2+4x^{4}z\frac{\mathrm{\partial <!--\; \partial \; -->}z}{\mathrm{\partial <!--\; \partial \; -->}x}=2x-<!--\; -\; -->4y^{5}z-<!--\; -\; -->4x{y}^5\frac{\mathrm{\partial <!--\; \partial \; -->}z}{\mathrm{\partial <!--\; \partial \; -->}x}$
Plugging inn $(1,-<!--\; -\; -->1,0)$ we get $\frac{\mathrm{\partial <!--\; \partial \; -->}z}{\mathrm{\partial <!--\; \partial \; -->}x}=\frac{1}{2}$.
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 $-<!--\; -\; -->4x{y}^{5}z-<!--\; -\; -->4x{y}^5\frac{\mathrm{\partial <!--\; \partial \; -->}z}{\mathrm{\partial <!--\; \partial \; -->}x}-<!--\; -\; -->20x{y}^{4}z\frac{\mathrm{\partial <!--\; \partial \; -->}y}{\mathrm{\partial <!--\; \partial \; -->}x}$.
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.

I have this practice problem before a test. Use implicit differentiation to find $dy/dx$ for the equation
$x^3+y^3=3xy.$
I have no idea how to do this, I didn't understand my lecturer. Can you guys show me the steps?

Let $r=r(t)$ and $\mathrm{\theta <!--\; \theta \; -->}=\mathrm{\theta <!--\; \theta \; -->}(t)$ with $r(t)>0$. Let $x(t)=r(t)\cos <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->}(t))$ and $y(t)=r(t)\sin <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->}(t))$ . Prove that $\frac{d\mathrm{\theta <!--\; \theta \; -->}}{dt}=\frac{1}{x^2+y^2}(x\frac{dy}{dt}-<!--\; -\; -->y\frac{dx}{dt})$
The hint is to use $y(t)/x(t)$ and use implicit differentiation but I can't see how to use that hint to solve this problem.

How do you find the second derivative of ${\displaystyle x^2{y}^2=1}$

Use implicit differentiation to find the points where the circle defined by x2+y2−2x−4y=−1 has horizontal tangent lines. List your answers as points in the form (a,b). 1. Find the points where the curve has a horizontal tangent line.
How do I solve this question? I know i have to solve for y' to find the gradient of the slope which I calculate to be y' = (2x-2)/(2y-4) y' = x-1/y-2
What do I do after this?

How do you differentiate ${\displaystyle x{e}^{x^2+y^2}}$

I want to solve $dy/dx$ for the following:
$x^2+y^2=R^2$ where $R$ is a constant.
I know to use implicit differentiation, though I have a question. When I derive $R^2$, do I obtain $2R$ or 0?
Additionally, deriving $y^2$ with respect to x yields $2y(dy/dx)$? This is different from a partial derivative?
Thanks!

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

How do you implicitly differentiate ${\displaystyle x^2+2xy-y^2+x=2}$

How do you find the derivative of ${\displaystyle x^{2}y}$

So I'm trying to differentiate an equation using implicit differentiation.
I start with $e^{x/y}=7x-<!--\; -\; -->y$
Now the left side of the eqn is where I'm having trouble.
I tried to use differentiation rules for exponentials, but this is incorrect.
Here's what I tried though:
$(e^x{)}^{1/y}\ln <!--\; \; -->(e^x)y^{\prime }=7-<!--\; -\; -->y^{\prime }$
simplfied to:
$x(e^x{)}^{1/y}\cdot <!--\; \cdot \; -->y^{\prime }=RS$

So my problem is that I have to use implicit differentiation to find the derivative of $F(x,y)=e^{xy}-<!--\; -\; -->x$ when $F(x,y)=10$ and the equation of the tangent at the point $(1,\log <!--\; \; -->(11))$. So I tried to solve this using two ways:
The first way I used was rearranging the equation to $10+x=e^{xy}$ and then using $\ln$ to simply into $\ln <!--\; \; -->(10+x)=xy$ before using implicit differentiation. The result is
$\frac{dy}{dx}=\frac{1}{10x+x^2}-<!--\; -\; -->\frac{y}{x}$
and after substituting the point, I get approximately −0.950 as the slope.
The second way I used was just differentiating it straight away rather than rearranging and I get
$\frac{dy}{dx}=\frac{\frac{1}{e^{xy}}-<!--\; -\; -->y}{x}.$
But when I sub the point in I get approximately −0.688 as the slope. So I'm not sure if I've done something wrong when getting the derivatives or am I not allowed to rearrange the equation?

Considering s is implicit to function of $p$, given by $s^6-<!--\; -\; -->p^4=1$. For what s is it increasing and decreasing?
Well, I answeblack first like following: Calculating the first derivative using the implicit differentiation is $s^{\prime }=\frac{2p^3}{3s^5}$. $s$ is increasing where $s^{\prime }>0$ and decreasing where $s^{\prime }<0$.
Obviously, I answeblack too generally and I'm wondering how I better answer it using the first derivative to support my answer.

There is no trouble when I have to differentiate a function $f(x,y)$ where $y=g(x)$ using Implicit Function Theorem. It's easy to get $g_{xx}$ and $g_{yy}$. For instance, the formula for $g_{xx}=-<!--\; -\; -->\frac{f_{xx}{F}_y^2-<!--\; -\; -->(F_{xy}+F_{yx})F_x{F}_y+F_{yy}{F}_x^2}{F_y^3}$ knowing that $g^{\prime }(x)=-<!--\; -\; -->\frac{F_x}{F_y}$ whenever $F_y\ne <!--\; \ne \; -->0$.
Now, i'm in the situation $z+x+(y+z{)}^4=0$ where it defines an implicit function $z=f(x,y)$.
I got $f_x$ and $f_y$ using the theorem. However, I can't get $f_{xx}$ and $f_{yy}$ from there. I know I shoud apply the chain rule but I get lost whenever trying to get a "general formula" for them.

The question asks to implicitly differentiate $\tan <!--\; \; -->(x+y)=x$.
I get the correct answer of $-<!--\; -\; -->{\sin }^2<!--\; \; -->(x+y)$. However, the book also says "or you can use $-<!--\; -\; -->{\displaystyle \frac{x^2}{x^2+1}}$". I don't see how I can construct a proper right triangle to give/obtain this equivalent answer. Can you assist?

When should I use total differentials and when should I use implicit differentiation?
An example of an equation where I'm unsure if to use total differentials or implicit differentiation is $f(x,y)=x^{2}y$
Using total differentials
$\frac{\mathrm{\partial <!--\; \partial \; -->}f}{\mathrm{\partial <!--\; \partial \; -->}x}dx+\frac{\mathrm{\partial <!--\; \partial \; -->}f}{\mathrm{\partial <!--\; \partial \; -->}x}dy=2xydx+x^{2}dy$, solving for $\frac{dy}{dx}=\frac{-<!--\; -\; -->2y}{x}$
Using implicit differentiation:
$(2x\frac{dy}{dx})(y)+x^2$, solving for $\frac{dy}{dx}=\frac{-<!--\; -\; -->x}{y}$
I'm unsure when to use each method and they both yield different results.

I need help finding the implicit differentiation of $x^4+\sin <!--\; \; -->y=x^3{y}^2$ I especially need help finding ${\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}}$ for both sides. I'm having trouble with the product rule on the right side with ${\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}}$ of $x^3{y}^2$.