Implicit Differentiation Made Easy: Expert Advice

Recent questions in Implicit Differentiation

June Mejia 2022-08-16

$x^{y} = y^{x}$
find an expression for $d y / d x$ in terms of $y$ and $x$
for the first part is $y x^{y - 1}$ and for the second part i got $\ln (y) y^{x} d y / d x$
when solving for dy/dx i got $y x^{y - 1}) / (\ln (y) y^{x})$
Just was wondering if my work is correct?
Sorry for the bad formatting, Im not sure how to make it formatted nice.

Databasex3 2022-08-16

How do you do implicit differentiation with the quotient rule?

tuanazado 2022-08-16

So I got this classic implicit differentiation problem $y = x^{x}$
I've tried to tackle it by transforming $x^{x}$ to the equivalent form of $e^{(\ln x) x}$ :
$y = x^{x}$
$y = (e^{\ln x})^{x}$
$y = e^{(\ln x) x}$
and then proceeded to differentiate it in the following way:
$y^{'} = e^{(\ln x) x} \cdot \ln x \cdot \frac{1}{x}$

metodystap9 2022-08-15

I understand the method used in implicit differentiation, it's just an application of the chain rule. But why can you define $y$ as a function of $x$ ?
In this equation for example:
$x^{2} + y^{2} = 1$
$2 x + 2 y y^{'} = 0$
Why isn't it just this?:
$2 x + 2 y = 0$
Thank you in advance.

Flambergru 2022-08-14

What is the derivative of $y = 6 x y$

zabuheljz 2022-08-14

Use implicit differentiation to calculate $\frac{d y}{d x}$ for $x^{2} + x y - 6 = y^{3}$

vrteclh 2022-08-14

I need to find $y^{''}$ by implicit differentiation : $2 x^{3} + 5 y^{3} = 2$

Gorlandint 2022-08-14

With the equation $x^{2} + 2 x y - 3 y^{2} + 16 = 0$ , I need to find the coordinates of the points on the curve where $\frac{d y}{d x} = 0$ . I think I have correctly used implicit differentiation to get $\frac{d y}{d x} = \frac{- x - y}{x - 3 y}$ . And the multiplied by the denominator and subsituted zero to get, $0 = - x - y$ , however, I'm not sure where to go from here.
If anyone could help that would be great. Thanks.

joyoshibb 2022-08-14

I am trying to solve the problem: $x^{2} + x y + y^{3} = 0$ using implicit differentiation.
My workings:
$(1) \frac{d}{d x} [x^{2}] + \frac{d}{d x} [x y] + \frac{d}{d x} [y^{3}] = \frac{d}{d x} [0]$
$(2) 2 x + y + \frac{d y^{3}}{d y} \frac{d y}{d x} = 0$
$(3) 2 x + y + 3 y^{2} (\frac{d y}{d x}) = 0$
$(4) \frac{d y}{d x} = - \frac{2 x + y}{3 y^{2}}$
But the answer says it should be:
$(3) 2 x + y + \frac{d x y}{d y} \frac{d y}{d x} + 3 y^{2} (\frac{d y}{d x}) = 0$
$(4) 2 x + y + \frac{d y}{d x} (x + 3 y^{2}) = 0$
$(5) \frac{d y}{d x} = - \frac{2 x + y}{x + 3 y^{2}}$
Why?

Dorsheele0p 2022-08-13

Find the gradient of the curve $x \ln y - \frac{x}{y} = 2$ at (−1,1)
I've done something wrong as I got the gradient to be 0 when the answer in the back says −0.5
Can someone help me with this question?
edit:
I got: $1 + \frac{1}{y} \frac{d y}{d x} - \frac{y - x \frac{d y}{d x}}{y^{2}}$
$y^{2} + y \frac{d y}{d x} - y - x \frac{d y}{d x} = 0$
$\frac{d y}{d x} = \frac{y - y^{2}}{y - x}$
Substituting the values in did not get me the answer

Meossi91 2022-08-13

Let $G (x, y) = x^{2} y^{4} - 3 x^{4} y$ .
(i) Find the first partial derivatives $G_{x}$ and $G_{y}$ .
(ii) Using (i) above, find $\frac{d y}{d x}$ .
(iii) If $G (x, y) = 0$ , confirm your answer in part (ii) above, finding $\frac{d y}{d x}$ using implicit differentiation.

metodystap9 2022-08-13

I'm trying to calculate this differential equation using implicit differentiation for $\frac{d y}{d x}$ in terms of x and y:
$2 x^{3} y + 5 x + 6 y - s i n (π y) = 8$
What I have tried:
By differentiating with respect to x
$6 x^{2} y + \frac{d y}{d x} 2 x^{3} + 5 + \frac{d y}{d x} 6 - \frac{d y}{d x} π \cos (π y) = 0$
I'm not sure where to go from here.

granuliet1u 2022-08-12

What is the derivative of $y^{2} = x$ where $y = \frac{1}{2}$

Dorsheele0p 2022-08-12

How do you implicitly differentiate $x + x y - 2 x^{3} = 2$

Trystan Castaneda 2022-08-12

What is the derivative of 7xy?

kvasilw0 2022-08-12

How do you find the second derivative by implicit differentiation?

Ledexadvanips 2022-08-12

Find the derivative $d y / d x$ if $x^{2} + y^{2} = 7$ , in terms of $x$ and $y$ . We differentiate both sides of the equation using the chain rule, and the fact that $d x / d x = 1$ . We then obtain:
$2 x \frac{d x}{d x} + 2 y \frac{d y}{d x} = 0$ , that is $2 x + 2 y \frac{d y}{d x} = 0$
So my question is totally basic: what does $\frac{d x}{d x}$ stand for? What's it doing in there in the first place? I thought $x^{2}$ could be differentiated directly to $2 x$ , without any intermediate steps, and I'm just not seeing what $\frac{d x}{d x}$ is doing in there. Thanks for any tips.

Kade Rosales 2022-08-11

Question:
Assume that y is a function of x. Find $\frac{d y}{d x}$ for $x^{3} + \cos^{3} (y) = 5 x$

My first move, using the chain rule, is:
$3 x^{2} + 3 \cos^{2} (y) (- \sin (y)) = 5$
However the text says this should instead be:
$3 x^{2} + 3 \cos^{2} (y) (- \sin (y)) \frac{d y}{d x} = 5$
I don't understand where or how the $\frac{d y}{d x}$ is being introduced.
To my mind, $\cos^{3} (y)$ = $(\cos (y))^{3}$ and so could be interpreted as:
$f (x) = x^{3}$
$g (x) = \cos (y)$
$F (x) = f (g (x)) = \cos^{3} (y)$
Then, using the Chain Rule:
$F^{'} (x) = f^{'} (g (x)) \cdot g^{'} (x) = 3 \cos^{2} (y) (- \sin (y))$
But apparently that's wrong.

sublimnes9 2022-08-11

I have some problem with implicit differentiation.
$y \cos x = 4 x^{2} + 3 y^{2}$
I am now stuck at the following equality and I don't know how to proceed. Can someone help me?
$y (- \sin x) + (\cos x) y^{'} = 8 x + 6 y y^{'}$

crazygbyo 2022-08-11

I am given the equation: $x^{a} y^{b} = 6$
Using implicit differentiation I find that the derivative of the equation with respect to y gives: $\frac{d}{d x} (y) = - \frac{a y}{b x}$ . However when I attempt to differentiate the "regular" way I don`t seem to reach the same answer. I would greatly appreciate someone walking me through the problem.

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