Recent questions in Implicit Differentiation
heelallev5 2022-08-05

Given:$\left[{\mathrm{tan}}^{-1}\left(x\right){\right]}^{2}+\left[{\mathrm{cot}}^{-1}\left(y\right){\right]}^{2}=1$Find the tangent line equation to the graph at the point (1,0) by implicit differentiationI found the derivative:$\frac{dy}{dx}=\frac{4{\mathrm{tan}}^{-1}\left(x\right)\cdot {\mathrm{cot}}^{-1}\left(y\right)}{\left({y}^{2}+1\right)\left({x}^{2}+1\right)}$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.$g\left(x\right)=3{x}^{-3}\left({x}^{4}-3{x}^{3}+15x-4\right)$

Tirimwb 2022-07-26

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

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$\frac{dy}{dx}=\frac{132{x}^{5}-132{x}^{32}y}{4{x}^{33}+7{y}^{6}}$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$y-y1=m\left(x-x1\right)$getting$y-1=0\left(x-1\right)$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 $\frac{{d}^{2}y}{{dx}^{2}}$ 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 ${y}^{\prime }$, the problem is:$\left(x-2y{\right)}^{3}=2{y}^{2}-3$So far I've been able to get this far:$3\left(x-2y{\right)}^{2}\left(1-2{y}^{\prime }\right)=4y\left({y}^{\prime }\right)$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 $dy/dx$ added?As clarification an example I will provide an example:Implicitly differentiate ${y}^{2}=x$.You get$2y\frac{dy}{dx}=1$as one of the first steps in differentiation. Why is the $dy/dx$ added after ${y}^{2}$ is differentiated?

Taniya Burns 2022-07-22

Find,${x}^{2}{y}^{2}-2xy+x=2$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 $d/dx$ and then ignore $dy/dx$ 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 ${f}^{\prime }\left(x\right)$

smuklica8i 2022-07-22

use implicit differentiation to find $\frac{dy}{dx}$ in terms of $x$ and $y$1.) ${x}^{3}-xy+{y}^{2}=4$2.) $y=\mathrm{sin}\left(xy\right)$find $\frac{{d}^{2}y}{d{x}^{2}}$3.) ${x}^{2}{y}^{2}-2x=3$

kislotd 2022-07-22

Find $\frac{dy}{dx}$ 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:$\frac{1-3{x}^{2}-2xy}{{x}^{2}+1}$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.:$-\frac{x\left(y+x{\right)}^{2}-y}{x}$Can anyone explain to me why I get different solutions ?

Israel Hale 2022-07-22

Given $y=y\left(x\right)$ what is$\frac{d}{dx}\left({e}^{x}\left({x}^{2}+{y}^{2}\right)\right)$I think its the $\frac{d}{dx}$ confusing me, I don't what effect it has compablack to $\frac{dy}{dx}$. Any help will be greatly appreciated.

Matilda Fox 2022-07-22

$x\mathrm{sin}\left(4x+5y\right)=y\mathrm{cos}\left(x\right)$I am trying to use implicit differentiation to find dx/dy for this problem but the answer i keep getting is$4x\mathrm{cos}\left(4x+5y\right)=-y\mathrm{sin}\left(x\right)$and I am stuck.

Lorelei Patterson 2022-07-22

I was wondering whether anybody could explain how you derive this implicit differentiation rule:$\frac{\mathrm{\partial }z}{\mathrm{\partial }x}=\frac{-\mathrm{\partial }f/\mathrm{\partial }x}{\mathrm{\partial }f/\mathrm{\partial }z}$if you have a function $z$ implicity defined by $f\left(x,y,z\right)=0$?

Zoagliaj 2022-07-21

The equation $2{x}^{4}{z}^{2}+{y}^{4}={x}^{2}-4x{y}^{5}z$ defines $z$ implicitly as a function of $x$ and $y$. Find $\frac{\mathrm{\partial }z}{\mathrm{\partial }x}$ at the point $\left(1,-1,0\right)$The solution was as follows:So if we derive with respect to $x$ we get$8{x}^{3}{z}^{2}+4{x}^{4}z\frac{\mathrm{\partial }z}{\mathrm{\partial }x}=2x-4{y}^{5}z-4x{y}^{5}\frac{\mathrm{\partial }z}{\mathrm{\partial }x}$Plugging inn $\left(1,-1,0\right)$ we get $\frac{\mathrm{\partial }z}{\mathrm{\partial }x}=\frac{1}{2}$.But when we differentiate with respect to $x$, on the last expression they used the chain rule on $x$ and $z$ and treated $y$ as a constant. Shouldn't you also use the chain rule on $y$ and get something like $-4x{y}^{5}z-4x{y}^{5}\frac{\mathrm{\partial }z}{\mathrm{\partial }x}-20x{y}^{4}z\frac{\mathrm{\partial }y}{\mathrm{\partial }x}$.I realize now that since $z=0$ in our point, the last expression would actually fall away and we would get the right answer. But the solution doesn't contain the last expression at all, so I'm confused about whether or not I've misunderstood implicit differentiation.

Makena Preston 2022-07-21

I have this practice problem before a test. Use implicit differentiation to find $dy/dx$ for the equation${x}^{3}+{y}^{3}=3xy.$I have no idea how to do this, I didn't understand my lecturer. Can you guys show me the steps?

Arectemieryf0 2022-07-21

Let $r=r\left(t\right)$ and $\theta =\theta \left(t\right)$ with $r\left(t\right)>0$. Let $x\left(t\right)=r\left(t\right)\mathrm{cos}\left(\theta \left(t\right)\right)$ and $y\left(t\right)=r\left(t\right)\mathrm{sin}\left(\theta \left(t\right)\right)$ . Prove that $\frac{d\theta }{dt}=\frac{1}{{x}^{2}+{y}^{2}}\left(x\frac{dy}{dt}-y\frac{dx}{dt}\right)$The hint is to use $y\left(t\right)/x\left(t\right)$ and use implicit differentiation but I can't see how to use that hint to solve this problem.

Luz Stokes 2022-07-20

How do you find the second derivative of ${x}^{2}{y}^{2}=1$

Paxton Hoffman 2022-07-20

Use implicit differentiation to find the points where the circle defined by x2+y2−2x−4y=−1 has horizontal tangent lines. List your answers as points in the form (a,b). 1. Find the points where the curve has a horizontal tangent line.How do I solve this question? I know i have to solve for y' to find the gradient of the slope which I calculate to be y' = (2x-2)/(2y-4) y' = x-1/y-2What do I do after this?

Matias Aguirre 2022-07-19