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## Find dy/dx by implicit differentiation if : x cos y=y?

turksvids
Featured 1 month ago

$\frac{\mathrm{dy}}{\mathrm{dx}} = \cos \frac{y}{1 + x \cdot \sin \left(y\right)}$

#### Explanation:

Take the derivative of both sides with respect to $x$:

$\frac{d}{\mathrm{dx}} \left(x \cdot \cos \left(y\right)\right) = \frac{d}{\mathrm{dx}} \left(y\right)$

Use the product rule on the left side:

$x \cdot - \sin \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} + \cos \left(y\right) = \frac{\mathrm{dy}}{\mathrm{dx}}$

Rearrange so all the terms with $\frac{\mathrm{dy}}{\mathrm{dx}}$ are on the same side:

$\frac{\mathrm{dy}}{\mathrm{dx}} + x \cdot \sin \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = \cos \left(y\right)$

Factor a little:

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(1 + x \cdot \sin \left(y\right)\right) = \cos \left(y\right)$

Solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \cos \frac{y}{1 + x \cdot \sin \left(y\right)}$

## How to solve the following ??

Eric Sia
Featured 1 month ago

$t = 50 \ln 10$

#### Explanation:

Given:

$k \frac{\mathrm{dV}}{\mathrm{dt}} + V = 0$

Rearrange:

$\frac{1}{V} \frac{\mathrm{dV}}{\mathrm{dt}} = - \frac{1}{k}$

Take the integral of both sides:

$\int \frac{1}{V} \mathrm{dV} = - \frac{1}{k} \int \mathrm{dt}$

This gives:

$\ln V = - \frac{t}{k} + C$

At $t = 0$, $V = {V}_{0}$:

$\ln {V}_{0} = C$

Hence:

$V = {V}_{0} {e}^{- \frac{t}{k}}$

Or:

$t = k \ln \left({V}_{0} / V\right)$

When $V = \frac{1}{10} {V}_{0}$:

$t = k \ln 10$

Given $k = 50$:

$t = 50 \ln 10$

## For what values of x is f(x)= -2x^3 + 4 x^2 + 5x -5  concave or convex?

Somebody N.
Featured 1 month ago

Concave: $\left\{x \in \mathbb{R} : \frac{2}{3} < x < \infty\right\}$

Convex: $\left\{x \in \mathbb{R} : - \infty < x < \frac{2}{3}\right\}$

#### Explanation:

A function is convex if the second derivative is positive and concave where its second derivative is negative. When the second derivative is $0$ this could mean the function is concave , convex, or it could be a point of inflexion. This would have to be tested using the first derivative.

$f \left(x\right) = - 2 {x}^{3} + 4 {x}^{2} + 5 x - 5$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(- 2 {x}^{3} + 4 {x}^{2} + 5 x - 5\right) = - 6 {x}^{2} + 8 x + 5$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} \left(- 6 {x}^{2} + 8 x + 5\right) = - 12 x + 8$

So:

$- 12 x + 8 < 0$

$x > \frac{8}{12}$

$x > \frac{2}{3}$

Concave: $\left\{x \in \mathbb{R} : \frac{2}{3} < x < \infty\right\}$

$- 12 x + 8 > 0$

$x < \frac{8}{12}$

$x < \frac{2}{3}$

Convex: $\left\{x \in \mathbb{R} : - \infty < x < \frac{2}{3}\right\}$

Graph:

graph{y=-2x^3+4x^2+5x-5 [-4, 4, -7.12, 7.12]}

## How to solve f'(x) = 0 for f(x)= ln(x) - e^(cos x) over 0<x<=10?

Eric Sia
Featured 1 month ago

Use Newton-Raphson method for numerical approximation.

#### Explanation:

$f \left(x\right) = \ln x - {e}^{\cos x}$

Take the derivative with respect to $x$:

$f ' \left(x\right) = \frac{1}{x} + \sin x {e}^{\cos x}$

Given the condition $f ' \left(x\right) = 0$:

$0 = \frac{1}{x} + \sin x {e}^{\cos x}$

Let

$u \left(x\right) = \frac{1}{x} + \sin x {e}^{\cos x}$

Plot $u \left(x\right)$ over $0 < x \le 10$:
graph{1/x+sinxe^(cosx) [-0.5, 10.5, -2, 4]}

By examination, the roots of $u \left(x\right)$ are close to $4$, $6$, and $10$.

Take the derivative with respect to $x$:

$u ' \left(x\right) = - \frac{1}{x} ^ 2 + \left(\cos x - {\sin}^{2} x\right) {e}^{\cos x}$

The Newton-Raphson approximation equation is:

${x}_{n + 1} = {x}_{n} - \frac{u \left({x}_{n}\right)}{u ' \left({x}_{n}\right)}$

Taking ${x}_{0} = 4$ gives:

${x}_{0} = 4$
${x}_{1} = 3.794906369$
${x}_{2} = 3.776053960$
${x}_{3} = 3.775920734$
${x}_{4} = 3.775920727$
${x}_{5} = 3.775920727$

Taking ${x}_{0} = 6$ gives:

${x}_{0} = 6$
${x}_{1} = 6.247412504$
${x}_{2} = 6.223989036$
${x}_{3} = 6.223939693$
${x}_{4} = 6.223939692$
${x}_{5} = 6.223939692$

Taking ${x}_{0} = 10$ gives:

${x}_{0} = 10$
${x}_{1} = 9.730092594$
${x}_{2} = 9.698501451$
${x}_{3} = 9.698236088$
${x}_{4} = 9.698236070$
${x}_{5} = 9.698236070$

Hence

$x = 3.775920727 , 6.223939692 , 9.698236070$

## What is the derivative of  y = ((x^2 + 4) / (x^2 - 4))^3?

Michael F.
Featured 2 weeks ago

The first derivative is $\quad \frac{- 48 x \cdot {\left({x}^{2} + 4\right)}^{2}}{{\left({x}^{2} - 4\right)}^{4}}$.

#### Explanation:

Let $\quad f \left(x\right) = {\left(\frac{{x}^{2} + 4}{{x}^{2} - 4}\right)}^{3}$.

Rearranging the term of $f$ gives us

$\quad f \left(x\right) = {\left(\frac{{x}^{2} + 4}{{x}^{2} - 4}\right)}^{3} = {\left({x}^{2} + 4\right)}^{3} / {\left({x}^{2} - 4\right)}^{3}$.

For differentiating we use the $\textcolor{g r e e n}{\text{chain rule}}$ for $\textcolor{b l u e}{{u}^{'} \left(x\right)}$ and $\textcolor{b l u e}{{v}^{'} \left(x\right)}$

$\textcolor{g r e e n}{\frac{d}{\mathrm{dx}} \left(u \left(v \left(x\right)\right)\right) = {\left(u \left(v \left(x\right)\right)\right)}^{'} = {u}^{'} \left(v \left(x\right)\right) \cdot {v}^{'} \left(x\right)}$ (outside times inside derivative)

and the $\textcolor{b l u e}{\text{quotient rule}}$ for the whole term of $f$

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \frac{u \left(x\right)}{v \left(x\right)} = {\left(\frac{u \left(x\right)}{v \left(x\right)}\right)}^{'} = \frac{{u}^{'} \left(x\right) \cdot v \left(x\right) - u \left(x\right) \cdot {v}^{'} \left(x\right)}{v \left(x\right)} ^ 2}$ (with $v \left(x\right) \ne 0$).

${f}^{'} \left(x\right) =$

$\frac{3 \cdot 2 x \cdot {\left({x}^{2} + 4\right)}^{2} \cdot {\left({x}^{2} - 4\right)}^{3} - {\left({x}^{2} + 4\right)}^{3} \cdot 3 \cdot 2 x \cdot {\left({x}^{2} - 4\right)}^{2}}{{x}^{2} - 4} ^ 6$.

Factorizing ${\left({x}^{2} - 4\right)}^{2}$ from both summands of the enumerator gives us

${f}^{'} \left(x\right) =$

$\frac{\cancel{{\left({x}^{2} - 4\right)}^{2}} \cdot \left[6 x \cdot {\left({x}^{2} + 4\right)}^{2} \cdot \left({x}^{2} - 4\right) - 6 x \cdot {\left({x}^{2} + 4\right)}^{3}\right]}{{x}^{2} - 4} ^ \cancel{{6}^{4}}$.

Factorizing $6 x \cdot {\left({x}^{2} + 4\right)}^{2}$ from both summands of the enumerator leads to

${f}^{'} \left(x\right) = \frac{6 x \cdot {\left({x}^{2} + 4\right)}^{2} \cdot \left[{x}^{2} - 4 - \left({x}^{2} + 4\right)\right]}{{x}^{2} - 4} ^ 4$

${f}^{'} \left(x\right) = \frac{6 x \cdot {\left({x}^{2} + 4\right)}^{2} \cdot \left(- 8\right)}{{x}^{2} - 4} ^ 4$.

The end result is

$\textcolor{red}{{f}^{'} \left(x\right) = \frac{- 48 x \cdot {\left({x}^{2} + 4\right)}^{2}}{{x}^{2} - 4} ^ 4}$.

I hope it helps.

## How to derive 5.18 equation? details in image

Truong-Son N.
Featured 1 week ago

LEFT-HAND INTEGRAL

The left-hand integral is trivial, as there are a lot of constants:

$\frac{\epsilon {A}_{s , r} \sigma}{\rho V c} {\int}_{0}^{t} \mathrm{dt} = \frac{\epsilon {A}_{s , r} \sigma}{\rho V c} \cdot t$

RIGHT-HAND INTEGRAL: PREPARATION

The right-hand integral starts as:

${\int}_{{T}_{i}}^{T} \frac{1}{{T}_{s u r}^{4} - T {'}^{4}} \mathrm{dT} '$

We wrote $T '$ here because it need not be equal to $T$. This is clearly a difference of quartics, so one can factor as follows:

$\frac{1}{{T}_{s u r}^{4} - T {'}^{4}} = \frac{1}{\left({T}_{s u r}^{2} + T {'}^{2}\right) \left({T}_{s u r}^{2} - T {'}^{2}\right)}$

$= \frac{1}{\left({T}_{s u r}^{2} + T {'}^{2}\right) \left({T}_{s u r} + T '\right) \left({T}_{s u r} - T '\right)}$

This becomes a partial fraction decomposition. We propose, at a constant surrounding temperature, the following two integrands:

$\frac{1}{\left({T}_{s u r}^{2} + T {'}^{2}\right) \left({T}_{s u r} + T '\right) \left({T}_{s u r} - T '\right)}$

$= {\overbrace{\frac{A T ' + B}{\left({T}_{s u r}^{2} + T {'}^{2}\right)}}}^{\text{Integrand 1" + overbrace(C/(T_(sur) + T') + D/(T_(sur) - T'))^"Integrand 2}}$

RIGHT-HAND INTEGRAL: ACQUIRING SYSTEM OF EQNS

When getting common denominators, one obtains:

$1 = \left(A T ' + B\right) \left({T}_{s u r}^{2} - T {'}^{2}\right) + C \left({T}_{s u r}^{2} + T {'}^{2}\right) \left({T}_{s u r} - T '\right) + D \left({T}_{s u r}^{2} + T {'}^{2}\right) \left({T}_{s u r} + T '\right)$

Next, simplify by distributing and re-factoring.

$1 = \textcolor{red}{A {T}_{s u r}^{2} T '} - \textcolor{\mathmr{and} a n \ge}{A T {'}^{3}} + \textcolor{p u r p \le}{B {T}_{s u r}^{2}} - \textcolor{c y a n}{B T {'}^{2}}$

$+ \textcolor{p u r p \le}{C {T}_{s u r}^{3}} - \textcolor{red}{C {T}_{s u r}^{2} T '} + \textcolor{c y a n}{C {T}_{s u r} T {'}^{2}} - \textcolor{\mathmr{and} a n \ge}{C T {'}^{3}}$

$+ \textcolor{p u r p \le}{D {T}_{s u r}^{3}} + \textcolor{red}{D {T}_{s u r}^{2} T '} + \textcolor{c y a n}{D {T}_{s u r} T {'}^{2}} + \textcolor{\mathmr{and} a n \ge}{D T {'}^{3}}$

This regroups into a polynomial in orders of $T '$ up to $3$:

$0 T {'}^{3} + 0 T {'}^{2} + 0 T ' + 1$

$= \textcolor{\mathmr{and} a n \ge}{\left(- A - C + D\right)} T {'}^{3} + \textcolor{c y a n}{\left(- B + C {T}_{s u r} + D {T}_{s u r}\right)} T {'}^{2} + \textcolor{red}{\left(A {T}_{s u r}^{2} - C {T}_{s u r}^{2} + D {T}_{s u r}^{2}\right)} T ' + \textcolor{p u r p \le}{\left(B {T}_{s u r}^{2} + C {T}_{s u r}^{3} + D {T}_{s u r}^{3}\right)}$

As a result, we have the following system of equations (since ${T}_{s u r} \ne 0$, we divided it out of $\left(3\right)$):

$- A - C + D = 0$ $\text{ "" "" "" "" } \boldsymbol{\left(1\right)}$
$- B + C {T}_{s u r} + D {T}_{s u r} = 0$ $\text{ "" } \boldsymbol{\left(2\right)}$
$A - C + D = 0$ $\text{ "" "" "" "" "" } \boldsymbol{\left(3\right)}$
$B {T}_{s u r}^{2} + C {T}_{s u r}^{3} + D {T}_{s u r}^{3} = 1$ $\text{ } \boldsymbol{\left(4\right)}$

RIGHT-HAND INTEGRAL: SOLVING SYSTEM OF EQNS

In solving this we could do the following.

• Since $- A - C + D = A - C + D$, it follows that $A = - A$, which is only true if $\textcolor{g r e e n}{A = 0}$.
• Therefore, from $\left(1\right)$ or $\left(3\right)$, $C = D$, and we can solve for $B$ in $\left(2\right)$.

$\implies B = 2 C {T}_{s u r} = 2 D {T}_{s u r}$

• To solve for $C$ or $D$, use $\left(4\right)$:

$2 C {T}_{s u r} \cdot {T}_{s u r}^{2} + C {T}_{s u r}^{3} + C {T}_{s u r}^{3} = 1$

$\implies \textcolor{g r e e n}{C = \frac{1}{4 {T}_{s u r}^{3}} = D}$

• Therefore, $\textcolor{g r e e n}{B} = 2 \cdot \frac{1}{4 {T}_{s u r}^{3}} \cdot {T}_{s u r} = \textcolor{g r e e n}{\frac{1}{2 {T}_{s u r}^{2}}}$.

RIGHT-HAND INTEGRAL: SETTING IT UP

This means we can insert $A$, $B$, $C$, and $D$ to get:

$\frac{1}{\left({T}_{s u r}^{2} + T {'}^{2}\right) \left({T}_{s u r} + T '\right) \left({T}_{s u r} - T '\right)}$

$= \frac{\textcolor{g r e e n}{\left(1 / 2 {T}_{s u r}^{2}\right)}}{\left({T}_{s u r}^{2} + T {'}^{2}\right)} + \frac{\textcolor{g r e e n}{\left(1 / 4 {T}_{s u r}^{3}\right)}}{{T}_{s u r} + T '} + \frac{\textcolor{g r e e n}{\left(1 / 4 {T}_{s u r}^{3}\right)}}{{T}_{s u r} - T '}$

$= {\overbrace{\frac{1}{2} \frac{1}{T} _ {\left(s u r\right)}^{2} \frac{1}{\left({T}_{s u r}^{2} + T {'}^{2}\right)}}}^{\text{Integrand 1" + overbrace(1/4 1/T_(sur)^3 [1/(T_(sur) + T') - 1/(T_(sur) - T')])^"Integrand 2}}$

The integration is now of the following:

${\int}_{{T}_{i}}^{T} \frac{1}{{T}_{s u r}^{4} - T {'}^{4}} \mathrm{dT} '$

$= {\overbrace{\frac{1}{2} \frac{1}{T} _ {\left(s u r\right)}^{2} {\int}_{{T}_{i}}^{T} \frac{1}{{T}_{s u r}^{2} + T {'}^{2}} \mathrm{dT} '}}^{\text{Integral 1" + overbrace(1/4 1/T_(sur)^3 int_(T_i)^(T) [1/(T_(sur) + T') - 1/(T_(sur) - T')]dT')^"Integral 2}}$

RIGHT-HAND INTEGRAL: SOLVING IT

The second integral here is (before evaluating the integral bounds):

$\textcolor{h i g h l i g h t}{\frac{1}{4} \frac{1}{T} _ {\left(s u r\right)}^{3} \int \left[\frac{1}{{T}_{s u r} + T '} - \frac{1}{{T}_{s u r} - T '}\right] \mathrm{dT} '}$

$= \frac{1}{4 {T}_{s u r}^{3}} \left[\ln | {T}_{s u r} + T ' | - \ln | {T}_{s u r} - T ' |\right]$

$= \textcolor{h i g h l i g h t}{\frac{1}{4 {T}_{s u r}^{3}} \ln | \frac{{T}_{s u r} + T '}{{T}_{s u r} - T '} |}$

The first integral here needs to be manipulated more. Factor out a ${T}_{s u r}^{2}$ to get:

$\textcolor{\mathrm{da} r k b l u e}{\frac{1}{2 {T}_{s u r}^{2}} \int \frac{1}{{T}_{s u r}^{2} + T {'}^{2}} \mathrm{dT} '}$

$= \frac{1}{2 {T}_{s u r}^{4}} \int \frac{1}{1 + {\left(T ' / {T}_{s u r}\right)}^{2}} \mathrm{dT} '$

This resembles the integral of $\arctan u$. Let $u = T ' / {T}_{s u r}$, so that $\mathrm{du} = \mathrm{dT} ' / {T}_{s u r}$.

Thus, ${T}_{s u r} \mathrm{du} = \mathrm{dT} '$, and that then gives us the cubed term on the outside like we expected when we integrate:

$\implies \frac{1}{2 {T}_{s u r}^{3}} \int \frac{1}{1 + {u}^{2}} \mathrm{du}$

$= \frac{1}{2 {T}_{s u r}^{3}} \arctan u$

$= \textcolor{\mathrm{da} r k b l u e}{\frac{1}{2 {T}_{s u r}^{3}} \arctan \left(\frac{T '}{T} _ \left(s u r\right)\right)}$

FORMING/SIMPLIFYING THE RESULT

Now we can combine the two overarching integrals to get:

${\int}_{{T}_{i}}^{T} \frac{1}{{T}_{s u r}^{4} - T {'}^{4}} \mathrm{dT} '$

$= | \left[\textcolor{h i g h l i g h t}{\frac{1}{4 {T}_{s u r}^{3}} \ln | \frac{{T}_{s u r} + T '}{{T}_{s u r} - T '} |} + \textcolor{\mathrm{da} r k b l u e}{\frac{1}{2 {T}_{s u r}^{3}} \arctan \left(\frac{T '}{T} _ \left(s u r\right)\right)}\right] {|}_{{T}_{i}}^{T}$

Evaluating this on the integral bounds, we get:

$= \left[\textcolor{h i g h l i g h t}{\frac{1}{4 {T}_{s u r}^{3}} \ln | \frac{{T}_{s u r} + T}{{T}_{s u r} - T} |} + \textcolor{\mathrm{da} r k b l u e}{\frac{1}{2 {T}_{s u r}^{3}} \arctan \left(\frac{T}{T} _ \left(s u r\right)\right)}\right] - \left[\textcolor{h i g h l i g h t}{\frac{1}{4 {T}_{s u r}^{3}} \ln | \frac{{T}_{s u r} + {T}_{i}}{{T}_{s u r} - {T}_{i}} |} + \textcolor{\mathrm{da} r k b l u e}{\frac{1}{2 {T}_{s u r}^{3}} \arctan \left(\frac{{T}_{i}}{T} _ \left(s u r\right)\right)}\right]$

Regroup terms to get:

$= \textcolor{h i g h l i g h t}{\frac{1}{4 {T}_{s u r}^{3}} \ln | \frac{{T}_{s u r} + T}{{T}_{s u r} - T} | - \frac{1}{4 {T}_{s u r}^{3}} \ln | \frac{{T}_{s u r} + {T}_{i}}{{T}_{s u r} - {T}_{i}} |} + \textcolor{\mathrm{da} r k b l u e}{\frac{1}{2 {T}_{s u r}^{3}} \arctan \left(\frac{T}{T} _ \left(s u r\right)\right) - \frac{1}{2 {T}_{s u r}^{3}} \arctan \left(\frac{{T}_{i}}{T} _ \left(s u r\right)\right)}$

To factor this, multiply the third and fourth terms by $\frac{2}{2}$, then factor out $\frac{1}{4 {T}_{s u r}^{3}}$ to get:

$= \frac{1}{4 {T}_{s u r}^{3}} \left\{\textcolor{h i g h l i g h t}{\ln | \frac{{T}_{s u r} + T}{{T}_{s u r} - T} | - \ln | \frac{{T}_{s u r} + {T}_{i}}{{T}_{s u r} - {T}_{i}} |} + \textcolor{\mathrm{da} r k b l u e}{2 \left[\arctan \left(\frac{T}{T} _ \left(s u r\right)\right) - \arctan \left(\frac{{T}_{i}}{T} _ \left(s u r\right)\right)\right]}\right\}$

FINALIZING THE RESULT

Now, we can equate this entire right-hand integral with the left-hand result we got on the first few lines:

$\frac{\epsilon {A}_{s , r} \sigma}{\rho V c} \cdot t$

$= \frac{1}{4 {T}_{s u r}^{3}} \left\{\ln | \frac{{T}_{s u r} + T}{{T}_{s u r} - T} | - \ln | \frac{{T}_{s u r} + {T}_{i}}{{T}_{s u r} - {T}_{i}} | + 2 \left[\arctan \left(\frac{T}{T} _ \left(s u r\right)\right) - \arctan \left(\frac{{T}_{i}}{T} _ \left(s u r\right)\right)\right]\right\}$

Multiply the left-hand constants over to get:

$\textcolor{b l u e}{t = \frac{\rho V c}{4 \epsilon {A}_{s , r} \sigma {T}_{s u r}^{3}} \left\{\ln | \frac{{T}_{s u r} + T}{{T}_{s u r} - T} | - \ln | \frac{{T}_{s u r} + {T}_{i}}{{T}_{s u r} - {T}_{i}} | + 2 \left[\arctan \left(\frac{T}{T} _ \left(s u r\right)\right) - \arctan \left(\frac{{T}_{i}}{T} _ \left(s u r\right)\right)\right]\right\}}$

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