Implicit differentiation Definition and 278 Threads

This is a text book example- i noted that we may have a different way of doing it hence my post. Alternative approach (using implicit differentiation); ##\dfrac{x}{y}=t## on substituting on ##y=t^2## we get, ##y^3-x^2=0## ##3y^2\dfrac{dy}{dx}-2x=0## ##\dfrac{dy}{dx}=\dfrac{2x}{3y^2}##...

My take; ##6x^2+6y+6x\dfrac{dy}{dx}-6y\dfrac{dy}{dx}=0## ##\dfrac{dy}{dx}=\dfrac{-6x^2-6y}{6x-6y}## ##\dfrac{dy}{dx}=\dfrac{-x^2-y}{x-y}## Now considering the line ##y=x##, for the curve to be parallel to this line then it means that their gradients are the same at the point##(1,1)##...

The question and ms guide is pretty much clear to me. I am attempting to use a non-implicit approach. ##\tan y=x, ⇒y=\tan^{-1} x## We know that ##1+ \tan^2 x= \sec^2 x## ##\dfrac{dx}{dy}=sec^2 y## ##\dfrac{dx}{dy}=1+\tan^2 y## ##\dfrac{dy}{dx}=\dfrac{1}{1+x^2}##...

If Tl;dr I am struggling in Math 171 and Physics 191 and throwing around the idea of declaring a geology major with an astronomy minor because the Physics major "juice is not worth the squeeze" at my age(29) anyone else out there who struggled with Calculus 1 when they first took it?Hello...

Hi PF A personal translation of a quote from Spanish "Calculus", by Robert A. Adams: It's about advice on Lebniz's notation1=(sec2⁡y)dydx means dxdx=(sec2⁡y)dydx, I'm quite sure. Why (sec2⁡y)dydx=(1+tan2⁡y)dydx? But I'm also quite sure that the right notation for (sec2⁡y)dydx=(1+tan2⁡y)dydx...

Hi, PF ##y^2=x## is not a function, but it is possible to obtain the slope at any point ##(x,y)## of the equation without previously clearing ##y^2##. It's enough to differentiate respect to ##x## the two members, treat ##y## like a ##x## differentiable function and make use of the Chain Rule...

I am confused about implicit differenciation in a few ways. The main confusion is why, in the equation ## x^2 + y^2 = 1 ##, when we are taking the derivative of the left side, ## 2x + 2yy\prime ##, are we adding a ## y\prime ## to the 2y but we aren't adding an ## x\prime ## to the 2x? I also...

The working I've tried is in the attachment.

Let $$Y(t)=tanh(ln(1+Z(t)^2))$$ where Z is the Hardy Z function; I'm trying to calculate the pedal coordinates of the curve defined by $$L = \{ (t (u), s (u)) : {Re} (Y (t (u) + i s (u)))_{} = 0 \}$$ and $$H = \{ (t (u), s (u)) : {Im} (Y (t (u) + i s (u)))_{} = 0 \}$$ , and for that I need to...

$\tiny{166.2.6.5}$ Find y' $$x^2-4xy+y^2=4$$ dy/dx $$2x-4(y+xy')+2yy'=2x-4y-4xy'+2yy'=0$$ factor $$y'(-4x+2y)=-2x+4y=$$ isolate $$y'=\dfrac{-2x+4y}{-4x+2y} =\dfrac{-x+2y}{-2x+y}$$ typo maybe not sure if sure if factoring out 4 helped

Summary:: van der waals I have a pretty good understanding of implicit differentiation. However I'm stuck on a homework problem and could use some help. [P + (an^2)/V^2][V - nb] = nRT a,n,b,R are constants My professor wants me to take the implicit differentiation of P wrt...

I am new to calculus. I am doing well in my class. I just have a few questions about implicit differentiation. First, why do we call it "implicit" differentiation? Also, when we do it, why when we differentiate a term with a "y" in it, why do we have to multiply it by a dY/dX? What does that...

Homework Statement: Let ##\frac{1}{a}=\frac{1}{b}+\frac{1}{c}## If ##\frac{db}{dt}=0.2## ,## \frac{dc}{dt}=0.3## , Find ##\frac{da}{dt}## when a=80 , b=100 Homework Equations: - Since we are supposed to find ##\frac{da}{dt}##, I can deduce that: ## \frac{da}{dt}...

Help! Keep running this and getting different answers, and none are right. 2xy^8 + 7xy = 27 at the point (3,1)

Okay so I'm really not sure where I went wrong here; here's how I worked through it: $$\ln\left(y+x\right)=x$$ $$\frac{\frac{dy}{dx}+1}{y+x}=1$$ $$\frac{dy}{dx}+1=y+x$$ If ##\ln\left(y+x\right)=x## then ##y+x=e^x## and ##y=e^x-x## $$\frac{dy}{dx}=y+x-1$$ $$\frac{dy}{dx}=e^x-x+x-1$$...

Homework Statement Find the value of h'(0) if: $$h(x)+xcos(h(x))=x^2+3x+2/π$$ Homework Equations Chain Rule Product Rule The Attempt at a Solution I differentiated both sides, giving h'(x)+cos(h(x))-xh'(x)sin(h(x))=2x+3 Next I factored out and isolated h'(x) giving me...

Homework Statement Find the equation of the tangent line to the graph of the given equation at the indicated point. ##xy^2+sin(πy)-2x^2=10## at point ##(2,-3)## Homework EquationsThe Attempt at a Solution Please see attached image so you can see my thought process. I think it would make more...

Homework Statement If ##x\sqrt{1+y} + y\sqrt{1+x } = 0##, then prove that ##\frac {dy} {dx} = \frac {-1}{(x-1)^2}##. 2.Relevant Equations: $$ \frac {dy} {dx} = - \frac {\left (\frac {\partial f}{\partial x} \right)} {\left( \frac {\partial f} {\partial y} \right)}.$$ 3...

Hello. My problem is as follows: Suppose x^4+y^2+y-3=0. a) Compute dy/dx by implicit differentiation. b) What is dy/dx when x=1 and y=1? c) Solve for y in terms of x (by the quadratic formula) and compute dy/dx directly. Compare with your answer in part a). I solved a) and b). a)=-4x^3/2y+1, and...

I need urgent help. I have this question: Use implicit differentiation to find an equation of the tangent line to the curve at the given point. \begin{equation} {x}^{2/3}+{y}^{2/3}=4 \\ \left(-3\sqrt{3}, 1\right)\end{equation} (astroid) x^{\frac{2}{3}}+y^{\frac{2}{3}}=4 My answer is...

The following lines of codes implements 1D diffusion equation on 10 m long rod with fixed temperature at right boundary and right boundary temperature varying with time. xsize = 10; % Model size, m xnum = 10; % Number of nodes xstp =...

Homework Statement Hello I have this circle with the equation : [/B] (x-a)^2+(y-b)^2=r^2 I want to find dy/dx for it 2. Homework Equations (x-a)^2+(y-b)^2=r^2 The Attempt at a Solution I am looking on the internet and it appears that I should use what is called "Implicit differentiation"...

For implicit differentiation, is dy/dx of x2+y2 = 50 the same as y2 = 50 - x2 ? From what I can take it, it'd be a no since. For x2+y2 = 50, d/dx (x2+y2) = d/dx (50) --- will eventually be ---> dy/dx = -x/y Where, y2 = 50 - x2 y = sqrt(50 - x2) dy/dx = .5(-x2+50)-.5*(-2x)

(a) Differentiate both sides of the equation with respect to x: $\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \left[ y^3 + y + x\,y^2 \right] &= \frac{\mathrm{d}}{\mathrm{d}x} \left[ 10 + 4\sin{(x)} \right] \\ 3\,y^2\,\frac{\mathrm{d}y}{\mathrm{d}x} +...

How do I figure whether to do implicit differentiation or just explicit?? Thanks for the answer.

Homework Statement Find the equation of the tangent line to the curve ##\ xy^2 + \frac 2 y = 4## at the point (2,1). Answer says ##\ y-1 = -\frac 1 2(x-2)## And with implicit differentiation I should have gotten ##\frac {dy} {dx}= -\frac {y^2} {2xy-\frac {2} {y^2}}## Homework Equations ##\...

Homework Statement \dfrac{x^2}{x+y}=y^2+8 Homework Equations Quotient Rule: \dfrac{g(x)\cdot f'(x)-g'(x)\cdot f(x)}{(g(x))^2} Product Rule: f(x)\cdot g'(x)+g(x)\cdot f'(x) The Attempt at a Solution \dfrac{(x+y\cdot\dfrac{dy}{dx})(2x)-(1\cdot\dfrac{dy}{dx})(x^2)}{(x+y\cdot...

Homework Statement For the given function z to demonstrate the equality: [/B]As you see I show the solution provided by the book, but I have some questions on this. I don't understand how the partial derivative of z respect to x or y has been calculated. Do you think this is correct? I...

Hey, I found a thread about part of what I'm trying to ask long ago: https://www.physicsforums.com/threads/implicit-differentiation.178328/ Basically, I noticed that if you multiply by x or by y in an equation before implicitly deriving, you get two different answers. Unfortunately their whole...

Homework Statement I am told to find dy/dx by implicit differentiation where: e^(x^2 * y) = x + y Homework Equations The above equation and the ln of it.The Attempt at a Solution e^(x^2 * y) = x + y (x^2 * y)ln(e) = ln(x+y) x^2 * y = ln(x+y) x^2(dy/dx) + y(2x) = 1/(x+y) * (1 + dy/dx)...

Here is the question: This is the step I came to after taking the derivatives and doing some simplification: ^ I did the work myself on paper, I just couldn't type out the whole thing clearly so that anyone else can see what I'm referring too... so I used some online tool to show that...

Homework Statement Find y'' Homework Equations 9x^2 +y^2 = 9 The Attempt at a Solution y' 18x+2y(y')=0 y'=-18x/2y y'=9x/y For the second derivative, I get the correct answer (same as the book) up until the very last step. Here's where I'm left at: -9( (-9x^2 - y^2) / y^3 ) The book then...

<< Mentor Note -- thread moved from the technical math forums at OP request, so no Homework Help Template is shown >>[/color] x2y + xy2 = 6 I know we use the chain rule from here, so wouldn't that be: (d/dx)(x2y + xy2) = (d/dx)(6) so using the chain rule of g'(x)f'(g(x) and the d/dx...

Homework Statement Find y' ... X^2+y^2=25I understand (I think) implicit differentiation, but there is one issue which hangs me up. I've done this before and this is just a refresher as my last calculus course was four years ago. From what I understand, 2x+2y(y')=0 But why isn't it...

First of all thanks for the help, i have a problem finding a good explanation of de ecuation (d/dx)f=(∂f/∂x)+(∂f/∂y)*(dy/dx) could anyone write me a good explanation of this ecuation? thanks for the help

What is ##\frac{d}{dx}(\frac{x}{y^2})##? Please tell me is it correct or not: ##\frac{d}{dx}(\frac{x}{y^2}) = \frac{[\frac{d}{dx}(x)] ⋅ (y^2) - (x) ⋅ [\frac{d}{dx} (y^2)]}{(y^2)^2}## ## = \frac{(x) ⋅ (y^2) - (x) ⋅ (\frac{d}{dy} (y^2)) ⋅ \frac{dy}{dx}}{y^4}## ##= \frac{xy^2 -...

Since we have this relationship between x and y, as the two sides are equal, so are their derivatives. We just have to remember that as y is a function of x, any function of y is also a function of x, with the inner function "y" composed inside whatever is being told to do to the y. So to...

To perform implicit differentiation we must make use of the chain rule. Basically if you have a function composed in another function, its derivative is the product of the inner function's derivative and the outer function's derivative. All other rules (such as the sum rule, the product rule...

Hello! Can someone help me with the process of solving $$\sqrt{x}$$+$$\sqrt{y}$$=5 on point (4,9)? With implicit, I differntiated both sides and ended up with 1/2x^-1/2+1/2y^-1/2$$\d{y}{x}$$=0 and I tried to isolate the dy/dx, but how do I get rid of the others? And with explicit, I isolated y...

So it has been quite a few years since I learned about implicit differentiation so the content is a bit rusty in my mind. x=rcos(θ) How do you find dx/dt? I know the answer but I am trying to figure out why. I mean dx/dt can be written as (dx/dθ)*(dθ/dt) so why is the answer not just...

Hi! I recently came upon this problem : the height of a right angled triangle is increasing at a rate of 5cm/min while the area is constant. How fast must the base be decreasing at the moment when the height is 5 times the base? I drew a picture of the triangle, labelled the height (h) and...

If $y^2+\cos(x+y) = 1$ find $\frac{dy}{dx}$. How do I differentiate $\cos(x+y)$ bit?

Mod note: Moved from the Homework section 1. Homework Statement This might seem like a stupid question but I'm unsure what z= ƒ(x/y) means? I'm not sure how I would find ∂z/∂x , ∂z/∂y just from this statement either. Thank you Homework EquationsThe Attempt at a Solution

Homework Statement Homework Equations The Attempt at a Solution Note: by real solution I mean the correct implicit derivative, not an actual real solution... Please help![/B]

Folks, Differentiate implicitly \phi(x,y)=0 I get: wrt to x \phi_x+\phi_y \frac{dy}{dx} and wrt to y \phi_y+\phi_x \frac{dx}{dy} however I don't know how this is derived \phi_x dx+\phi_y dy=0

Hi Folks, It is been given that differentiation of \phi(x,y)=0 is \phi_{x} dx+ \phi_{y} dy=0 however I arrive at \phi_{x} dx/dy+ \phi_{y} dy/dx=0 via the chain rule. Where \phi_{x}=d \phi/dx etc What am I doing wrong? Thanks

I have an equation: r^2 = x^2 So I found out dr/dx = x/r. But when I try to find the second derivative, I get d2r/dx2 = -x^2/r^3 when the text says it should be (r^2 - x^2)/r^3. Can anyone help? My working out: r^2 - x^2 = 0 r^2 = x^2. Assume r is a function of x. rr' = x (first derivative...

Homework Statement with answers given: Homework Equations use implicit differentiation The Attempt at a Solution I always get this answer but not the second one PLs explain the second answer for I am very desperate. Thank You

$$6x-\sqrt{2xy}+xy^3 ={y}^{2}$$ $$6-?+3x{y}^{2}{y'}^{}+{y}^{3}=2y{y'}^{}$$ Got stumped on this one answer was complicated...

I'm having some trouble with the terminology used in calculus. My book states: "Fortunately we don't need to solve an equation for Y in terms of X in order to find the derivative of Y. Instead we can use the method of implicit differentiation. This consists of differentiating both sides of the...