Implicit Differentiation Made Easy: Expert Advice

Recent questions in Implicit Differentiation

iroroPagbublh 2022-08-11

Find $d y / d x$ through implicit differentiation:
$\sin (x) = x (1 + \tan (y))$
My solution
$\begin{aligned} (1) & \frac{d}{d x} [\sin (x)] & = \frac{d}{d x} [x (1 + \tan (y))] \\ (2) & \cos (x) & = (1) (1 + \tan (y)) + x (1 + \tan (y))^{- 1} (\sec^{2} (y)) \frac{d y}{d x} \\ (3) & \cos (x) & = (1 + \tan (y)) + \frac{x (\sec^{2} (y))}{1 + \tan (y)} \frac{d y}{d x} \\ (4) & \cos (x) - (1 + \tan (y)) & = \frac{x (\sec^{2} (y))}{1 + \tan (y)} \frac{d y}{d x} \\ (5) & \frac{d y}{d x} & = \frac{(\cos (x) - 1 - \tan (y)) (1 + \tan (y))}{x (\sec^{2} (y))} \end{aligned}$
Solution from manual I'm using
$\begin{aligned} (6) & \sin (x) & = x (1 + \tan (y)) \\ (7) & \cos (x) & = x (s e c^{2} (y)) y^{'} + (1 + \tan (y)) (1) \\ (8) & y^{'} & = \frac{\cos (x) - \tan (y) - 1}{x \sec^{2} (y)} \end{aligned}$

June Mejia 2022-08-11

How do you differentiate $e^{x y} + x^{2} - y^{2} = 0$

Meossi91 2022-08-11

I am trying to implicitly differentiate the following function:
$λ = \exp [{(α + \frac{s}{λ - s})}^{- 1}]$
Can someone help me with this?

Samson Kaufman 2022-08-11

I'm trying to compute the (implicit) derivative $\frac{d y}{d x}$ from the function
$e^{2 y} = x^{3}$
If I use $\ln$ on both sides I can isolate $y$ and find the derivative:
$\ln (e^{2 y}) = \ln (x^{3})$
$2 y = 3 \ln (x)$
$y = \frac{3}{2} \ln (x)$
$y^{'} = \frac{3}{2 x}$
But if I use implicit differentiation I get:
$\frac{d}{d x} (e^{2 y}) = \frac{d}{d x} x^{3}$
$\frac{d}{d y} (e^{2 y}) \frac{d}{d x} (y) = 3 x^{2}$
$2 e^{2 y} \cdot y^{'} = 3 x^{2}$
$y^{'} = \frac{3 x^{2}}{2 e^{2 y}}$
I know both methods should give the same result. What am I missing?

ljudskija7s 2022-08-10

As we know, for implicit differentiation we have $\frac{\partial y}{\partial x} = - \frac{F_{x}}{F_{y}}$
but if I use a trick that $\frac{\partial y}{\partial x} = \frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}} = \frac{\partial F}{\partial x} \frac{\partial y}{\partial F} = \frac{\partial y}{\partial x}$ as $\partial F$ cancels each other out, so it is like $\frac{\partial y}{\partial x} = \frac{F_{x}}{F_{y}}$ without the "minus", where did I go wrong?

targetepd 2022-08-10

$y = x - a \sin (b x) - 3$
where $x = x (a, b)$
We have to use implicit differentiation to calculate $\frac{d x}{d a}, \frac{d x}{d b}$ .
This is what I get.
$\frac{d y}{d a} = \frac{d x}{d a} - \sin (b x)$
for $\frac{d x}{d a}$ and
$\frac{d y}{d b} = \frac{d x}{d b} - a b \cos (b x) \frac{d x}{d b}$
for $\frac{d x}{d b}$
Can someone please let me know if I am doing it right. Thanks

Mark Elliott 2022-08-10

I am struggling with the following problem for implicit differentiation.
I am tasked to differentiate implicitly the following function, and evaluate $y^{″} (0)$ , where $y = y (x)$ .
$\ln (y + 1) + \sin (x y) = \ln (5) .$
I have differentiated this once to find,
$(y + x y^{'}) \cos (x y) + \frac{y^{'}}{y + 1} = 0$
But how to advance from here to find $y^{″} (0)$ ?

sunnypeach12 2022-08-09

The following differentiated implicitly with respect to θ:
$3 x = \tan θ$
The book says
$3 d x = \sec^{2} θ d θ$
One could start the calculation like this (I think):
$\frac{d}{d θ} 3 x = \frac{d}{d θ} \tan θ$
From there I'm not really sure about the steps.

sarahkobearab4 2022-08-09

How do you find the linear approximation of $f (x, y) = \sqrt{53 - 9 x^{2} - y^{2}}$ at (2,-1)?

heelallev5 2022-08-05

Given:
$[\tan^{- 1} (x)]^{2} + [\cot^{- 1} (y)]^{2} = 1$
Find the tangent line equation to the graph at the point (1,0) by implicit differentiation

I found the derivative:
$\frac{d y}{d x} = \frac{4 \tan^{- 1} (x) \cdot \cot^{- 1} (y)}{(y^{2} + 1) (x^{2} + 1)}$
I may have done my derivative wrong, but my main concern is at some point 0 will be plugged into cot inverse, resulting in division by zero.

Awainaideannagi 2022-07-28

Differentiate the function.
$g (x) = 3 x^{- 3} (x^{4} - 3 x^{3} + 15 x - 4)$

Tirimwb 2022-07-26

Differentiate.
$F (x) = \frac{1}{8 x - 5}$

PoentWeptgj 2022-07-23

I was looking to implicitly differentiate
$- 22 x^{6} + 4 x^{33} y + y^{7} = - 17$
and found it to be
$\frac{d y}{d x} = \frac{132 x^{5} - 132 x^{32} y}{4 x^{33} + 7 y^{6}}$
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
$y - y 1 = m (x - x 1)$
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?

Donna Flynn 2022-07-23

How do you use implicit differentiation to find $\frac{d^{2} y}{{d x}^{2}}$ of $x^{3} + y^{3} = 1$

Donna Flynn 2022-07-23

I'm stuck on this problem involving implicit differentiation.
The instructions ask me to find $y^{'}$ , the problem is:
$(x - 2 y)^{3} = 2 y^{2} - 3$
So far I've been able to get this far:
$3 (x - 2 y)^{2} (1 - 2 y^{'}) = 4 y (y^{'})$
I've been trying to manipulate it for a while but I can't figure out how to finish the problem properly.

Israel Hale 2022-07-23

Why, after differentiating $y$ on one side of the equation, is $d y / d x$ added?
As clarification an example I will provide an example:
Implicitly differentiate $y^{2} = x$ .
You get
$2 y \frac{d y}{d x} = 1$
as one of the first steps in differentiation. Why is the $d y / d x$ added after $y^{2}$ is differentiated?

Taniya Burns 2022-07-22

Find,
$x^{2} y^{2} - 2 x y + x = 2$
find the slope and the equation of the tangent line at (2,0) by implicit differentiation.
Part of the problem is that my text book says find $d / d x$ and then ignore $d y / d x$ after I get
$x^{2} y^{2} = 2 x^{2} y$
Please show me where I am going wrong. I would be able to find the equation of the tangent line if I had the correct $f^{'} (x)$

smuklica8i 2022-07-22

use implicit differentiation to find $\frac{d y}{d x}$ in terms of $x$ and $y$

1.) $x^{3} - x y + y^{2} = 4$

2.) $y = \sin (x y)$
find $\frac{d^{2} y}{d x^{2}}$

3.) $x^{2} y^{2} - 2 x = 3$

kislotd 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:
$\frac{(x + y) 2 x - (1 - \frac{d y}{d x}) x^{2}}{(x + y)^{2}} = 2 y \frac{d y}{d x}$
I blackuced that to get:
$2 x^{2} + 2 x y - x^{2} - x^{2} * d y / d x = (2 y * d y / d x) (x + y)^{2}$
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:
$d y / d x = (2 y (x + y)^{2}) / (4 x^{7} y)$
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!

skilpadw3 2022-07-22

Find $\frac{d y}{d x}$ for $x^{2} = \frac{x - y}{x + y}$
I have solved this in two ways.
First, I multiplicated the whole equation by $x + y$ and then I calculated the implicit derivative. I got the following solution:
$\frac{1 - 3 x^{2} - 2 x y}{x^{2} + 1}$
So far so good. When I calculated the implicit derivative of the original expression using the quotient rule though, I got a different solution, i.e.:
$- \frac{x (y + x)^{2} - y}{x}$
Can anyone explain to me why I get different solutions ?

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