Implicit Differentiation Made Easy: Expert Advice

Recent questions in Implicit Differentiation

Jean Farrell 2022-09-24

I have the following expression which I need to implicitly differentiate:
$x y^{2} + x^{2} + y + \sin (x^{2} y) = 0$
I'm a little confused as I'm not entirely sure what to do with the trig function. Here is my work so far:
$\frac{d y}{d x} [x y^{2} + x^{2} + y + \sin (x^{2} y)] = \frac{d y}{d x} 0$
$\frac{d y^{2}}{d x} + 2 x + \frac{d y}{d x} + \cos (x^{2} y) (2 x \frac{d y}{d x}) = 0$
How should I proceed?

Averi Fields 2022-09-23

When I search implicit differentiation for equation $x^{2} + y^{2} = r^{2}$ I find results of two versions: one using derivative and the other using differential.

Version1: $\frac{d}{d x} (x^{2} + y^{2} = r^{2}) \Leftrightarrow 2 x + 2 y \frac{d y}{d x} = 0$
Version2: $d (x^{2} + y^{2} = r^{2}) \Leftrightarrow 2 x d x + 2 y d y = 0$

Using both methods, I can derive the result: $\frac{d y}{d x} = - \frac{x}{y}$

However, I am confused, could you please provide some answers to:

1. Which one (derivative /differential) is the "real" implicit differentiation?
2. What are the differences between using these two methods?
3. When should differential be used rather than derivative?

Conrad Beltran 2022-09-23

Consider the implicit function
$(w (x) + 1) e^{w (x)} = x .$
I need to approximate $w (1.1)$ using the fact that $w (1) = 0$ .

Could you give me any hints?

Amaris Orr 2022-09-23

What is the derivative of $e^{\cos x} + e^{\sin y} = \frac{1}{4}$

mangicele4s 2022-09-22

If I am doing an implicit differentiation in a curve in $R^{2}$ for example, because I want to get the tangent line on a certain point... How am I suposed to know if I should differentiate it in terms of $Y$ , or in terms of $X$ ?
Am I going to get two slopes, one that's perpendicular to the other one by changing the term that I am doing that differentiation? If so, how can I guarantee that the term choosed to differentiate will give me the slope for the tangent?!

Hagman7v 2022-09-22

What is the slope of the curve at t=3 assuming that the equations define x and y implicitly as differentiable functions $x = f (t)$ , $y = g (t)$ , and $x = t^{5} + t$ , $y + 4 t^{5} = 4 x + t^{4}$ ?

sombereki51 2022-09-21

If $y + \sin (y) = \cos (x)$ , then
$\frac{d y}{d x} = - \frac{\sin (x)}{1 + \cos (y)},$
so $\frac{d y}{d x}$ is not defined when $\cos (y) = - 1.$

But $\cos (x)$ is differentiable at every real number, so $y + \sin (y)$ should also be differentiable, and so $y$ should be differentiable at every point. But it is not differentiable when ? $\cos (y) = - 1$

Urraiyg 2022-09-19

I have been given the following first order derivative equation:
$\frac{d x}{d t} = a x - b x y$
How do I find the second derivative? I know how to find implicit derivatives but since the differential in this case needs to be found in terms of $d x / d t$ , instead of the $d y / d x$ which I am used to, I am confused. Should I used the product rule to find the derivative of $- b x y$ ?

Krish Crosby 2022-09-18

$f (x, y) = x^{2} + y^{2} - 1$
$0 = x^{2} + y^{2} - 1 \Rightarrow y = \sqrt{1 - x^{2}}$
I differentiaded $g (x) = \sqrt{1 - x^{2}}$ with the chain rule and got $g^{'} (x) = - \frac{x}{\sqrt{1 - x^{2}}}$ .
Can someone tell me how to do it with implicit differentiation?
I tried this formula $y^{'} (x) = - \frac{f_{x}}{f_{y}} = - \frac{x}{y}$ , the solution should obviously be the same so I guess I might not be allowed to use the formula here? We had the implicit function theorem and $d F (x, y) (\begin{matrix} h \\ k \end{matrix}) = (\begin{matrix} h \\ d f (x, y) (h, k) \end{matrix})$ but I don't really know how to apply this here.

maredilunavy 2022-09-17

Problem:

Use implicit differentiation to compute $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ of the function $x^{3} + y^{3} + z^{3} - 3 x y z = 0$ .
What I Got:
$3 x^{2} + 0 + 3 z^{2} \cdot \frac{\partial z}{\partial x} - 3 y \cdot \frac{\partial z}{\partial x}$
$\frac{\partial z}{\partial x} (3 z^{2} - 3 y) = - 3 x^{2}$
$\frac{\partial z}{\partial x} = \frac{- 3 x^{2}}{3 z^{2} - 3 y}$
and
$0 + 3 y^{2} + 3 z^{2} \cdot \frac{\partial z}{\partial y} - 3 x \cdot \frac{\partial z}{\partial y}$
$\frac{\partial z}{\partial y} (3 z^{2} - 3 x) = - 3 y^{2}$
$\frac{\partial z}{\partial y} = \frac{- 3 y^{2}}{3 z^{2} - 3 y}$
For what I have got, is this the correct way to do this question?

videosfapaturqz 2022-09-16

How do you find dy/dx for the function $y = \sin (2 x + 4 y)$

Julia Chang 2022-09-16

My first one is express $\frac{d y}{d x}$ in terms of $x$ and $y$ if $x^{2} - 4 x y^{3} + 8 x^{2} y = 20$
what I did is this
$2 x - 4 x (3 y^{2}) (\frac{d y}{d x}) + 4 y^{3} + 8 x^{2} \frac{d y}{d x} + 16 x y = 0$
$- 4 x (3 y^{2} \frac{d y}{d x}) + 8 x^{2} \frac{d y}{d x} = - 2 x - 4 y^{3} - 16$
Finally I got $\frac{d y}{d x} = \frac{2 x - 4 y^{3} - 16}{- 4 (x) (3 y^{2}) + 8 x^{2}}$
My second question is
Find an equation for the tangent line to the graph of $2 x^{2} - 5 y^{2} + x y = 5$ at the point (2,1)
Anyway I simplified the equation to
$\frac{d y}{d x} = \frac{- 4 x - 1}{- 10 y + x}$ plugging in x and y I got $\frac{9}{8}$
so my line is $y - 1 = \frac{9}{8} (x - 2)$ but I am unsure if if I did this correctly.

Ivan Buckley 2022-09-15

What is the derivative of $e^{2 x y}$

Alexus Deleon 2022-09-15

Find partial derivative of $z$ with respect to the partial of $y$ using the result of the chain rule.
$\ln (x^{2} + y^{2}) + x \ln (z) - \cos (x y z) = 3.$
I would use regular implicit differentiation for this problem but what does "using result of the chain rule" mean?

batejavizb 2022-09-14

Find dydx by implicit differentiation.
x2−4xy+y2=4

My solution:
$\frac{d}{d x} x^{2} - \frac{d}{d x} 4 x y + \frac{d}{d x} y^{2} = \frac{d}{d x} 4$
$2 x - (4 x)^{'} (y) + (4 x) (y)^{'} + 2 y \frac{d y}{d x} = 0$
$2 x - 4 y \frac{d y}{d x} + 4 x \frac{d y}{d x} + 2 y \frac{d y}{d x} = 0$
$\frac{d y}{d x} (- 4 y + 4 x + 2 y) = - 2 x$
$\frac{d y}{d x} = \frac{- 2 x}{4 x - 2 y}$
Correct answer:
$\frac{d y}{d x} = \frac{2 y - x}{y - 2 x}$
I would like to understand why my answer is incorrect. If someone could take a look at my steps and explain to me where I went wrong it would be greatly appreciated!

Tiana Hill 2022-09-06

I'm stuck on this implicit differentiation question, where I'm requiblack to find the turning points, the equation goes like this $y^{3} + 3 x y^{2} - x^{3} = 3$

rancuri5a 2022-09-06

How do you find the derivative of y in the equation $\ln (x y) = x + y$ ?

Brynn Collins 2022-08-29

Suppose we have the equation $V = \frac{1}{3} π r^{2} h$ . Find $\frac{d r}{d h}$ .
[Chain Rule]
We have $\frac{d V}{d h} = \frac{d V}{d r} \cdot \frac{d r}{d h}$
${\begin{cases} \frac{d V}{d h} = \frac{1}{3} π r^{2} \\ \frac{d V}{d r} = \frac{2}{3} π r h \end{cases} ⟹ \frac{1}{3} π r^{2} = \frac{2}{3} π r h \cdot \frac{d r}{d h} ⟹ r = 2 h \cdot \frac{d r}{d h} ⟹ \frac{d r}{d h} = \frac{r}{2 h}$
[Implicit Differentiation]
$\begin{aligned} \frac{d}{d h} V & = \frac{π}{3} \frac{d}{d h} (r^{2} h) \\ 0 & = \frac{π}{3} (r^{2} \cdot \frac{d}{d h} h + h \frac{d}{d h} r^{2}) \\ 0 & = r^{2} + 2 \cdot h \cdot r \frac{d r}{d h} \\ \frac{d r}{d h} & = - \frac{r^{2}}{2 \cdot h \cdot r} = - \frac{r}{2 h} \end{aligned}$
Why does the solution using the chain rule method have a different sign compablack to the one using implicit differentiation? Did I make any mistake?

Eisregiony1 2022-08-29

$x^{y} = y^{x}$
is the equation
My solution is :
$\frac{d y}{d x} = (\ln y - \frac{y}{x}) / (\ln x - \frac{x}{y})$
Just wondering if this is correct!

roletatx 2022-08-28

$\ln (1 + x y) = x y$
When I try to implicitly differentiate this I get
$\frac{1}{1 + x y} (y + x y^{'})$ = (y + xy')
At which point the $(y + x y^{'})$ terms cancel out, leaving no $y^{'}$ to solve for.
However, the answer to this is $- \frac{y}{x}$ ... How do you get this?

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