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

How to determine dy/dx given ${\displaystyle x+\sin y=y+1}$?

Given that ${\displaystyle x^2-y^2=25}$ then show ${\displaystyle \frac{d^{2}y}{{dx}^2}=-\frac{25}{y^3}}$?

How to find the derivative of ${\displaystyle y=\sin (x+y)}$?

How to find the derivative of ${\displaystyle \sqrt{x^2+y^2}}$?

Help me to find dy/dx by implicit differentiation
$x{e}^y-<!--\; -\; -->15x+5y=0$

I'm struggling somewhat to understand how to use implicit differentiation to solve the following equation:
$\cos <!--\; \; -->\cos <!--\; \; -->(x^3{y}^2)-<!--\; -\; -->x\cot <!--\; \; -->y=-<!--\; -\; -->2y$
I figublack that the calculation requires the chain rule to differentiate the composite function, but I'm not sure how to 'remove' the y with respect to x from inside the composite function. My calculations are:
$\frac{dy}{dx}[\cos <!--\; \; -->\cos <!--\; \; -->(x^3{y}^2)-<!--\; -\; -->x\cot <!--\; \; -->y]=\frac{dy}{dx}[-<!--\; -\; -->2y]$
$\frac{dy}{dx}[\cos <!--\; \; -->\cos <!--\; \; -->(x^3{y}^2)]=\sin <!--\; \; -->\cos <!--\; \; -->(x^3{y}^2\cdot <!--\; \cdot \; -->y^{\prime }(x))\cdot <!--\; \cdot \; -->\sin <!--\; \; -->(x^3{y}^2\cdot <!--\; \cdot \; -->y^{\prime }(x))\cdot <!--\; \cdot \; -->6x^{2}y\cdot <!--\; \cdot \; -->y^{\prime }(x)$
This seems a bit long and convoluted. I'm also not sure how this will allow me to solve for $y^{\prime }(x)$. Carrying on...
$\frac{dy}{dx}[x\cot <!--\; \; -->y]=-<!--\; -\; -->{\csc }^2<!--\; \; -->y\cdot <!--\; \cdot \; -->y^{\prime }(x)$
$\frac{dy}{dx}[-<!--\; -\; -->2y]=-<!--\; -\; -->2$
Is my calculation correct so far? This seems to be a very complex derivative. Any comments or feedback would be appreciated.

I need to find $\frac{dy}{dx}$:
$\frac{d}{dx}{e}^y\cos <!--\; \; -->x=1+\sin <!--\; \; -->(xy)$
using implicit differentiation.
So far I've gotten to this point which I don't think I'm doing right because then I get stuck:
$e^y(-<!--\; -\; -->\sin <!--\; \; -->x)+(\cos <!--\; \; -->x)(e^y)(\frac{dy}{dx})=\cos <!--\; \; -->(xy)(x)(\frac{dy}{dx})+y(1)$
Am I right? Am I wrong? If I'm right I definitely have no clue where to go...and if I'm wrong...then again I have no idea what to do.

If z is defined implicitly as a function of x and y, find $\frac{\mathrm{\partial <!--\; \partial \; -->}z}{\mathrm{\partial <!--\; \partial \; -->}x}$
$yz=ln(x+z)$
I've attempted this equation going forward with implicit differentiation and I've used the theorem that states $\frac{\mathrm{\partial <!--\; \partial \; -->}z}{\mathrm{\partial <!--\; \partial \; -->}x}=-<!--\; -\; -->\frac{\mathrm{\partial <!--\; \partial \; -->}F/\mathrm{\partial <!--\; \partial \; -->}x}{\mathrm{\partial <!--\; \partial \; -->}F/\mathrm{\partial <!--\; \partial \; -->}z}$

Please excuse if the formatting of this post is wrong.

There's a question that asks for the 2nd derivative of $y-<!--\; -\; -->2x-<!--\; -\; -->3xy=2$

From what I know, I have to use implicit differentiation, using which I get:
$\frac{12+18y}{(1-<!--\; -\; -->3x{)}^2}$
But can you solve for y in the initial equation and differentiate two times (aka explicit differentiation)? By doing that I got:
$\frac{48}{(1-<!--\; -\; -->3x{)}^3}$
I'm not sure if this is a correct answer as I am new to differentiation. I guess that this question is also tied with another question; can you substitute y in implicit differentiation by solving for it in the initial equation?

I need to find the derivatives of
$(1+xy{)}^3=y{x}^{-<!--\; -\; -->1}$
I tried this and stuck
$3(1+xy{)}^2(0+x(\frac{dy}{dx})+y)=-<!--\; -\; -->y{x}^{-<!--\; -\; -->2}+(x^{-<!--\; -\; -->1}\frac{dy}{dx})$
$\frac{dy}{dx}(x+y-<!--\; -\; -->x^{-<!--\; -\; -->1})=-<!--\; -\; -->y{x}^{-<!--\; -\; -->2}-<!--\; -\; -->3(1+xy{)}^2$
$\frac{dy}{dx}=\frac{-<!--\; -\; -->y{x}^2-<!--\; -\; -->3(1+xy{)}^2}{x+y-<!--\; -\; -->x^{-<!--\; -\; -->1}}$
I'm not quite sure how to simplify it to achieve the final answer -
$\frac{-<!--\; -\; -->y(3x^2(1+xy{)}^2+1}{x(3x^2(1+xy{)}^2-<!--\; -\; -->1}$

What is the derivative of ${\displaystyle x=\tan (x+y)}$

When I check my answer using the some website, I get a result I can't agree with. So I'd like to hear your opinion :)
Asked: use implicit differentiation to differentiate $\sin <!--\; \; -->(xy)$.
My take on the matter:
Using the chain rule: $\frac{d}{dx}\sin <!--\; \; -->(xy)$ = $\cos <!--\; \; -->(xy)\frac{d}{dx}xy$, then using the product rule on latter factor we get: $\frac{d}{dx}xy=y+x\frac{dy}{dx}$
Hence: $\frac{d}{dx}\sin <!--\; \; -->(xy)=y\cos <!--\; \; -->(xy)+x\cos <!--\; \; -->(xy)\frac{dy}{dx}$.
Is this correct?

How do you differentiate ${\displaystyle x{y}^2-x^{3}y=6}$

4xy^3+5xy=27 Point:(3,1)
So, using Implicit Differentiation...
(4y^3+4xy^2(dy/dx))+(5y+5x(dy/dx))=0
4xy^2(dy/dx)+5x(dy/dx)=-4y^3-5y
dy/dx(4xy^2-5x)=-4y^3-5y
dy/dx=(-4y^3-5y)/(4xy^2+5x)
Now I replace x and y in the equation with x=3 and y=1
I get 20/27, which is apparently incorrect. What am I doing wrong?

I'm supposed to implicitly differentiate the following and give the answer in terms of $y\mathrm{\prime <!--\; \prime \; -->}$
$\tan <!--\; \; -->(x-<!--\; -\; -->y)=\frac{y}{1+x^2}$
$\frac{(1+x^2)y\mathrm{\prime <!--\; \prime \; -->}-<!--\; -\; -->2xy}{(1+x^2{)}^2}$
How do I solve for $y\mathrm{\prime <!--\; \prime \; -->}$?
Edit: After a ludicrous amount of algebra, i finally ended up at
$\frac{se{c}^2(x-<!--\; -\; -->y)(1+x^2{)}^2+2xy}{(1+x^2)(1+se{c}^2(x-<!--\; -\; -->y)(1+x^2))}$
which apparently is correct.

find $\frac{dy}{dx}$ of $x^3+y^3+8$ using partial derivative
So what I need to do is $\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}x}$ and $\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}y}$ which are:
$3x^2$ and $3y^2$ and than to divide to get $\frac{\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}x}}{\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}y}}=\frac{\mathrm{\partial <!--\; \partial \; -->}y}{\mathrm{\partial <!--\; \partial \; -->}x}=1$?

$x\log <!--\; \; -->(x)+y\log <!--\; \; -->(y)=1$
${\displaystyle \frac{dy}{dx}}=?$
I calculated $\frac{dy}{dx}=-<!--\; -\; -->\frac{1+\log <!--\; \; -->(x)}{1+\log <!--\; \; -->(y)}$
however, the correct answer seems to be $-<!--\; -\; -->\log <!--\; \; -->(x)/\log <!--\; \; -->(y)$
I'm confused, can someone help?

Using implicit differentiation, what is $\frac{dy}{dx}$ if $xy+4=x$?
I know the answer is $\frac{1-<!--\; -\; -->y}{x}$ but I don't know how to get the answer.
So, I believe it is using a product rule. However, do I also need to use the chain rule?
What I did is
$xy+4=x$
I got $x+y$ using the product rule so then,
$x\left(\frac{dy}{dx}\right)+y+0=1$
And is 4 a constant?
I am not sure the next step why $\frac{dy}{dx}$ belongs to $x$ instead of $y$? Can anyone please explains this to me?

For the circle $x^2+y^2=r^2$, show that $\mid \frac{y^{″}}{[1+(y^{\prime }{)}^2{]}^{3/2}}\mid$
Is $x$ a constant as it disappear from the final equation? Should I only try to differentiate $y$ and $r$?

Surfaces $x^2+y^2+z^2+w^2=1$ and $x+2y+3z+4w=2$ define implicitly functions $x=x(y,z)$ and $w=w(y,z)$. Show that $\frac{\mathrm{\delta <!--\; \delta \; -->}x}{\mathrm{\delta <!--\; \delta \; -->}y}=\frac{2x-<!--\; -\; -->4y}{4x-<!--\; -\; -->w}$ when $4x\ne <!--\; \ne \; -->w$

Attempt to solve
Now i am not quite sure what i am suppose to do here ? I suppose compute $\frac{\mathrm{\delta <!--\; \delta \; -->}x}{\mathrm{\delta <!--\; \delta \; -->}y}$ via implicit-differentiation since both of these surfaces are given in implicit form. Another problem is that i am suppose to form system of equations from two of these surfaces since $x=x(y,z)$ and $w=w(y,z)$ are referring to same function ? Something like this ?
$F(x(y,z),y,z,w(y,z))=\{\begin{array}{l}{x}^2+y^2+z^2+w^2=1\\ x+2y+3z+4w=2\end{array}$
I don't think i have very good understanding on what i am actually suppose to do here so if someone can hint me in right direction / maybe try to explain on what i am suppose to do. I don't expect someone to handover complete solutions.