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

Stephanie Hunter
2022-07-17
Answered

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

You can still ask an expert for help

Jorge Franklin

Answered 2022-07-18
Author has **11** answers

Use the product rule:

$d\frac{(u.v)}{dx}=u.\frac{dv}{dx}+v.\frac{du}{dx}$

$\left({x}^{2}y\right)\prime ={x}^{2}.\frac{dy}{dx}+y2x$

Or

$\left({x}^{2}y\right)\prime ={x}^{2}y\prime +2yx$

$d\frac{(u.v)}{dx}=u.\frac{dv}{dx}+v.\frac{du}{dx}$

$\left({x}^{2}y\right)\prime ={x}^{2}.\frac{dy}{dx}+y2x$

Or

$\left({x}^{2}y\right)\prime ={x}^{2}y\prime +2yx$

asked 2022-08-25

How do calculate Partial differentiation of implicit function

$f(x+y+z,{x}^{2}+{y}^{2}+{z}^{2})=0$

$f(x+y+z,{x}^{2}+{y}^{2}+{z}^{2})=0$

asked 2022-08-10

$y=x-a\mathrm{sin}(bx)-3$

where $x=x(a,b)$

We have to use implicit differentiation to calculate $\frac{dx}{da},\frac{dx}{db}$.

This is what I get.

$\frac{dy}{da}=\frac{dx}{da}-\mathrm{sin}(bx)$

for $\frac{dx}{da}$ and

$\frac{dy}{db}=\frac{dx}{db}-ab\mathrm{cos}(bx)\frac{dx}{db}$

for $\frac{dx}{db}$

Can someone please let me know if I am doing it right. Thanks

where $x=x(a,b)$

We have to use implicit differentiation to calculate $\frac{dx}{da},\frac{dx}{db}$.

This is what I get.

$\frac{dy}{da}=\frac{dx}{da}-\mathrm{sin}(bx)$

for $\frac{dx}{da}$ and

$\frac{dy}{db}=\frac{dx}{db}-ab\mathrm{cos}(bx)\frac{dx}{db}$

for $\frac{dx}{db}$

Can someone please let me know if I am doing it right. Thanks

asked 2022-08-12

Find the derivative $dy/dx$ 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 $dx/dx=1$. We then obtain:

$2x\frac{dx}{dx}+2y\frac{dy}{dx}=0$, that is $2x+2y\frac{dy}{dx}=0$

So my question is totally basic: what does $\frac{dx}{dx}$ stand for? What's it doing in there in the first place? I thought ${x}^{2}$ could be differentiated directly to $2x$, without any intermediate steps, and I'm just not seeing what $\frac{dx}{dx}$ is doing in there. Thanks for any tips.

$2x\frac{dx}{dx}+2y\frac{dy}{dx}=0$, that is $2x+2y\frac{dy}{dx}=0$

So my question is totally basic: what does $\frac{dx}{dx}$ stand for? What's it doing in there in the first place? I thought ${x}^{2}$ could be differentiated directly to $2x$, without any intermediate steps, and I'm just not seeing what $\frac{dx}{dx}$ is doing in there. Thanks for any tips.

asked 2022-11-16

What is the derivative of $x=\mathrm{tan}(x+y)$

asked 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?

${x}^{3}+{y}^{3}=3xy.$

I have no idea how to do this, I didn't understand my lecturer. Can you guys show me the steps?

asked 2022-08-14

I am trying to solve the problem: ${x}^{2}+xy+{y}^{3}=0$ using implicit differentiation.

My workings:

$(1)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{d}{dx}[{x}^{2}]\phantom{\rule{thinmathspace}{0ex}}+\frac{d}{dx}[xy]\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}\frac{d}{dx}[{y}^{3}]=\frac{d}{dx}[0]$

$(2)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2x+y+\frac{d{y}^{3}}{dy}\frac{dy}{dx}=0$

$(3)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2x+y+3{y}^{2}\phantom{\rule{thinmathspace}{0ex}}(\frac{dy}{dx})=0$

$(4)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{dy}{dx}=\overline{){\displaystyle -\frac{2x+y}{3{y}^{2}}}}$

But the answer says it should be:

$(3)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2x+y+\frac{dxy}{dy}\frac{dy}{dx}+3{y}^{2}\phantom{\rule{thinmathspace}{0ex}}(\frac{dy}{dx})=0$

$(4)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2x+y+\frac{dy}{dx}(x+3{y}^{2})=0$

$(5)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{dy}{dx}=-\frac{2x\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}y}{x\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}3{y}^{2}}$

Why?

My workings:

$(1)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{d}{dx}[{x}^{2}]\phantom{\rule{thinmathspace}{0ex}}+\frac{d}{dx}[xy]\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}\frac{d}{dx}[{y}^{3}]=\frac{d}{dx}[0]$

$(2)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2x+y+\frac{d{y}^{3}}{dy}\frac{dy}{dx}=0$

$(3)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2x+y+3{y}^{2}\phantom{\rule{thinmathspace}{0ex}}(\frac{dy}{dx})=0$

$(4)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{dy}{dx}=\overline{){\displaystyle -\frac{2x+y}{3{y}^{2}}}}$

But the answer says it should be:

$(3)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2x+y+\frac{dxy}{dy}\frac{dy}{dx}+3{y}^{2}\phantom{\rule{thinmathspace}{0ex}}(\frac{dy}{dx})=0$

$(4)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2x+y+\frac{dy}{dx}(x+3{y}^{2})=0$

$(5)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{dy}{dx}=-\frac{2x\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}y}{x\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}3{y}^{2}}$

Why?

asked 2022-09-30

I'm sure this question has been asked before and yet I can't find it.

Is there a proof of implicit differentiation or is it simply an application of the chain rule? If it's the former, could you give or point me to the proof? If the latter, could you explain exactly how conceptually it works (or point to a link that does so)?

Is there a proof of implicit differentiation or is it simply an application of the chain rule? If it's the former, could you give or point me to the proof? If the latter, could you explain exactly how conceptually it works (or point to a link that does so)?