Implicit Differentiation Made Easy: Expert Advice

Recent questions in Implicit Differentiation

onetreehillyg 2022-08-26

Given:
$e^{7 z} = x y z$
The task is to compute the partial derivatives dz/dx and dz/dy using implicit differentiation. My solution is now as follows:
$(d z / d x) 7 e^{7 z} = y z + (d z / d x) x y$
And so,
$(d z / d x) (7 e^{7 z} - x y) = y z$
Therefore correct answer is,
$\frac{d z}{d x} = \frac{y z}{(7 e^{7 z} - x y)}$

zibazeleor 2022-08-25

How do calculate Partial differentiation of implicit function
$f (x + y + z, x^{2} + y^{2} + z^{2}) = 0$

kweqiwaix 2022-08-25

Say you are given the equation
$p (V - b) = R T e^{- \frac{a}{R V T}}$
Where p, b, R and a are all constants.
And you are asked to find an expression for $\frac{d V}{d T}$ , would the left side be a chain rule with a product rule with an implicit differential?So far I believe that the right hand side differentiates to:
$\frac{a}{T V} e^{\frac{- a}{R V T}} + \frac{d V}{d T} * \frac{a}{V^{2}} e^{\frac{- 1}{R V T}}$
Is that fully correct?

rkus2zg0 2022-08-25

Find $\frac{d y}{d x}$ by implicit differentiation
$x^{2} - 4 x y + y^{2} = 4$
I know to take the derivatives of both sides, which would be:
$\frac{d}{d x} [x^{2} - 4 x y + y^{2}] = 0$
I'm not sure if I did it right, but I then got:
$2 x - 4 \cdot (x \frac{d y}{d x}) + y + 2 y (\frac{d y}{d x}) = 0$
I don't know where to go from here, or even if the previous step is correct. Please help!
Edit: I have followed the advice given and I ended up with:
$\frac{x - 2 y}{2 x - 1}$
However this was incorrect. Someone please tell me what I am missing here.

Kaydence Villegas 2022-08-22

I am currently working on a question which involves me differentiating
$\frac{y}{x}$
I can't find nothing in books or on the internet about how to deal with this kind of implicit differentiation.

jiwekeegi 2022-08-21

I am struggling with an Implicit Differentiation question which is as follows:

$z = (7 x^{4}) * \ln (x) 4$ where $z$ and $x$ are functions of $t$ . $\frac{d x}{d t} = 4$ when $x = e$ . Calculate $\frac{d z}{d t}$ .

What I have tried so far is finding $\frac{d z}{d x}$ which I believe is $(7 x^{3}) + (28 x^{3}) * \ln (x)$ which I have used the chain rule to get this answer for $\frac{d z}{d x}$ .

This is the part where I have tried several things such as multiplying $\frac{d z}{d x}$ by $\frac{d x}{d t}$ but I'm not sure how that works and I have tried replacing the $x$ 's with $e$ 's and equation $\frac{d x}{d t}$ to $\frac{d z}{d x}$ but I am really unsure what I'm supposed to be doing in this question and nothing appears right.

Any help would be greatly appreciated.

Rosemary Burns 2022-08-21

I am trying to find the derivative $\frac{d y}{d x}$ of the following function $y e^{x y} = \sin x$ using implicit differentiation.
What I have is the following using product rule:
$\frac{d}{d x} y e^{x y} = \frac{d}{d x} \sin x$
$⟹ y^{2} e^{x y} + \frac{d y}{d x} e^{x y} = \cos x$
However, I noticed in the solution manual it has it as the following and not sure why:
$\frac{d}{d x} y e^{x y} = \frac{d}{d x} \sin x$
$⟹ y e^{x y} (x \frac{d y}{d x} + y) + \frac{d y}{d x} e^{x y} = \cos x$

Felix Fitzgerald 2022-08-20

How would I do these problems using Implicit Differentiation? I don't understand how to do them. Help?
Find $d y / d x$ if $\cos (4 x) - 2 x e^{4 y} = 0$ .
and
Find slope of the tangent line to the curve $\sqrt{12 x + 8 y} + \sqrt{2 x y} = 12$ at the point $(4, 2)$ .

Hudson Bullock 2022-08-20

Find $y^{'}$ if $y = \sec^{- 1} (\sqrt{3 x})$ . I know that $\sec^{- 1} (x) = \frac{1}{x \sqrt{x^{2} - 1}}$ but if I plug in $\sqrt{3 x}$ do I use the product rule on the bottom along with the chain rule?
I realize there are easier ways, but I must use Implicit differentiation to solve this problem.

Expositur3e 2022-08-20

How do you differentiate $x^{2} + y^{2} = 2 x y$

Cyrus Munoz 2022-08-20

Use implicit differentiation to find the derivative of $\arcsin (y + x)$
I have no idea how to proceed. I now that the derivative of $\arcsin (x)$ is $\frac{1}{\sqrt{1 - x^{2}}}$ but I don't know how to incorporate that into this implicit differentiation problem.

podmitijuy0 2022-08-19

Find dy/dx by implicit differentiation. $3 x^{2} + 3 = \ln 5 x y^{2}$
This is what i did: $6 x = D (\ln 5 x + \ln 5 y^{2})$ But i was thinking i could also use this way:
$6 x = D (5 x y^{2}) / 5 x y^{2}$
Are both ways the same?

cottencintu 2022-08-19

I would like to find $\frac{d y}{d x}$ , where we have the implicit equation
$(2 x + y)^{4} + 3 x^{2} y = y^{3} .$
I just can't seem to understand the problem and which rule of calculus to use. Could someone please further elaborate?

Memphis Khan 2022-08-19

Let $z = z (x, y)$ be defined implicitly by $F (x, y, z (x, y)) = 0$ , where $F$ is a given function of three variables.
Prove that if $z (x, y)$ and $F$ are differentiable, then
$\frac{d z}{d x} = - \frac{\frac{d F}{d x}}{\frac{d F}{d z}}$
if $d F / d z \neq 0$
I am kind of confused with $z = z (x, y)$ . Should I differentiate $d z$ / $d x$ and $d z$ / $d y$ , but there's nothing inside $x$ and $y$ so what should I differentiate $d x$ and $d y$ with then? I've been told that this is something related to implicit differentiation. In some other examples, they do use differentiate $F$ i.e. $d F$ but in most examples, I only see that they are differentiating the variables inside it.

ljudskija7s 2022-08-18

I'm learning implicit differentiation and I've hit a snag with the following equation.
$f (x, y) = x + x y + y = 2$
$D x (x) + D x (x y) + D x (y) = D x (2)$
$1 + x y^{'} + y + y^{'} = 0$
$x y^{'} + y^{'} = - 1 - y$
$y^{'} (x + 1) = 1 + y$
$y^{'} = \frac{(1 + y)}{(x + 1)}$
$y^{″} = \frac{(x + 1) y^{'} - (1 + y)}{(x + 1)^{2}}$
$y^{″} = \frac{(x + 1) \frac{(1 + y)}{(x + 1)} - (1 + y)}{(x + 1)^{2}}$
Ok now what according to this y'' = 0 which is wrong.

odigavz 2022-08-18

How does implicit differentiation work?

Dillan Valenzuela 2022-08-17

My equation is $y = x^{y^{2}}$
I did the ln of both sides, then I tried implicit differentiation. I got
$y^{'} = \frac{x^{y^{2}} y^{2}}{x} .$

Yair Valentine 2022-08-17

If $4 x^{2} + 5 x + x y = 4$ and $y (4) = - 20$ , find $y^{'} (4)$ by implicit differentiation.
I implicitly differentiated the equation, but I don't see how I can use $y (4) = - 20$ to my advantage.

Sydney Stein 2022-08-16

How do you find the derivative of $x e^{y} = y - 1$

empalhaviyt 2022-08-16

Find $\frac{d y}{d x}$ by implicit differentiation. Note that $y$ is a function of $x$ .
Given:
$e^{\frac{x}{y}} = x - y$
My solution:
Take derivative of left-hand side
$\frac{d y}{d x} e^{\frac{x}{y}} = (e^{\frac{x}{y}}) (\frac{y (1) - (1) y^{'} x}{y^{2}}) = (e^{\frac{x}{y}}) (\frac{y - y^{'} x}{y^{2}})$
Take derivative of right-hand side
$\frac{d y}{d x} (x) - \frac{d y}{d x} (y) = 1 - 1 (y^{'}) = 1 - y^{'}$
The expression is now
$(e^{\frac{x}{y}}) (\frac{y - y^{'} x}{y^{2}}) = 1 - y^{'}$
Multiply both sides by $y^{2}$
$e^{\frac{x}{y}} y - y^{'} x = y^{2} - y^{'} y^{2}$
Rearrange variables
$y e^{\frac{x}{y}} - y^{2} = y^{'} x - y^{'} y^{2}$
Factor $y^{'}$
$y e^{\frac{x}{y}} - y^{2} = y^{'} (x - y^{2})$
Factor $(x - y^{2})$ out of right-hand side, divide on the left-hand side
$\frac{y e^{\frac{x}{y}} - y^{2}}{(x - y^{2})} = y^{'}$
The correct answer from the textbook is:
$\frac{y^{2} - y e^{\frac{x}{y}}}{y^{2} - x e^{\frac{x}{y}}} = y^{'}$
Where did I go wrong? Thanks for your help.

Implicit differentiation is a method used to find the derivative of an equation in which the equation is not written in the form "y = something". It involves differentiating each side of the equation with respect to the same variable, typically "x", and is useful for finding derivatives of equations with multiple variables and higher-order derivatives. Examples include finding derivatives of equations involving circles, lines, and parabolas. By practicing implicit differentiation, students can learn to solve complex equations and understand the relationship between a function and its derivative.

Calculus 2

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