difficult y11 diff qs? (2 Viewers)

Average Boreduser · Dec 19, 2023

WeiWeiMan said:
I never did implicit wth

its literally just chain rule dw.

Average Boreduser · Dec 19, 2023

WeiWeiMan said:
I may be coping hard for not knowing how to solve it but that’s way harder than anything that’s realistically gonna appear in the HSC

recall log laws simplify from there and you're golden my dude.

WeiWeiMan · Dec 19, 2023

Average Boreduser said:
its literally just chain rule dw.

Yeah ik but I never learnt it properly
Also ru accelerating 4u

Average Boreduser · Dec 19, 2023

WeiWeiMan said:
Yeah ik but I never learnt it properly
Also ru accelerating 4u

nah im just doing it to get ahead. im only acceling 3u (smh priv schools and their dumb rules fr)

WeiWeiMan · Dec 19, 2023

Average Boreduser said:
recall log laws simplify from there and you're golden my dude.

Too hard

WeiWeiMan · Dec 19, 2023

Average Boreduser said:
nah im just doing it to get ahead. im only acceling 3u (smh priv schools and their dumb rules fr)

Just negotiate with head of math or smth brotha

Average Boreduser · Dec 19, 2023

WeiWeiMan said:
Too hard

too unrealisting tbh lol. but not impossible. let y=ln........... simplify and you'll get smn easy to diff.

Aeonium · Dec 19, 2023

Luukas.2 said:
There is the evil differentiation question from Ascham (evil if you don't know what to do to work it out, I mean)...

$\text{Find } \frac{d}{dx}\left[\log_e{\left(\frac{x^2e^{x^3}}{\sqrt{x^2+1}}\right)}\right]$

$\begin{align*} \frac d{dx} \left[\ln\left(\frac{x^2e^{x^3}}{\sqrt{x^2+1}}\right)\right]&=\frac d{dx} \left(\ln (x^2e^{x^3})-\ln\sqrt{x^2+1}\right)\\ &=\frac{2xe^{x^3}+3x^4e^{x^3}}{x^2e^{x^3}}-\frac{\frac{2x}{\sqrt{x^2+1}}}{\sqrt{x^2+1}}\\ \end{align*}$
i cant be bothered simplifying - is this correct?

Average Boreduser · Dec 19, 2023

WeiWeiMan said:
Just negotiate with head of math or smth brotha

nah tbh I'm not tryna acc 4u bc im scared i might fk it if i do it too early lmao

Average Boreduser · Dec 19, 2023

Aeonium said:
$\begin{align*} \frac d{dx} \left[\ln\left(\frac{x^2e^{x^3}}{\sqrt{x^2+1}}\right)\right]&=\frac d{dx} \left(\ln (x^2e^{x^3})-\ln\sqrt{x^2+1}\right)\\ &=\frac{2xe^{x^3}+3x^4e^{x^3}}{x^2e^{x^3}}-\frac{\frac{2x}{\sqrt{x^2+1}}}{\sqrt{x^2+1}}\\ \end{align*}$
i cant be bothered simplifying - is this correct?

expand the ln(x^2e^..) and take 1/2 out of ln(x^2+1) from your d/dx

Aeonium · Dec 19, 2023

Average Boreduser said:
expand the ln(x^2e^..) and take 1/2 out of ln(x^2+1) from your d/dx

just felt like product rule LOL forgot the 2 on the denom rip it's this right $\frac{d}{dx}\sqrt{f(x)}=\frac{f'(x)}{2\sqrt{f(x)}$

Average Boreduser · Dec 19, 2023

Aeonium said:
just felt like product rule LOL forgot the 2 on the denom rip it's this right $\frac{d}{dx}\sqrt{f(x)}=\frac{f'(x)}{2\sqrt{f(x)}$

nah i meant u can expand further using ln(AB)=lnA+lnB and lnsqrtx=1/2lnx

Aeonium · Dec 19, 2023

Average Boreduser said:
nah i meant u can expand further using ln(AB)=lnA+lnB and lnsqrtx=1/2lnx

oh yeah fair ngl i neglect some laws sometimes

WeiWeiMan · Dec 19, 2023

Average Boreduser said:
nah tbh I'm not tryna acc 4u bc im scared i might fk it if i do it too early lmao

4u is way harder than 3u ig

Average Boreduser · Dec 19, 2023

WeiWeiMan said:
4u is way harder than 3u ig

exactly hence why im not gonna try and persuade my teacher bc i think id be extra prepped if i waited the yr

WeiWeiMan · Dec 19, 2023

Average Boreduser said:
exactly hence why im not gonna try and persuade my teacher bc i think id be extra prepped if i waited the yr

personally idk what i'd do for 4u if i hadn't alry done 3u lol

Luukas.2 · Dec 19, 2023

$\begin{align*} y &= x^k + (c - x)^k \\ \frac{dy}{dx} &= kx^{k - 1} + k(c - x)^{k - 1} \times -1 = k\left[x^{k - 1} - (c - x)^{k - 1}\right] \\ \\ \text{Put $\frac{dy}{dx} = 0$:} \quad k\left[x^{k - 1} - (c - x)^{k - 1}\right] &= 0 \\ \left(\frac{x}{c - x}\right)^{k - 1} &= 1 \qquad \text{as $k \neq 0$} \\ x &= \frac{c}{2} \\ \text{Put $x = \frac{c}{2}$:} \quad y &= \left(\frac{c}{2}\right)^k +\left(c - \frac{c}{2}\right)^k = 2\left(\frac{c}{2}\right)^k \\ \\ \text{Thus, the stationary point is at} \quad &\left(\frac{c}{2},\ 2\left(\frac{c}{2}\right)^k\right) \\ \\ \frac{d^2y}{dx^2} &= k(k - 1)x^{k - 2} - k(k - 1)(c - x)^{k - 2} \times -1 \\ &= k(k - 1)\left[x^{k - 2} + (c - x)^{k - 2}\right] \\ \\ \text{Put $x = \frac{c}{2}$:} \quad \frac{d^2y}{dx^2} &= k(k - 1)\left[\left(\frac{c}{2}\right)^{k - 2} + \left(c - \frac{c}{2}\right)^{k - 2}\right] = 2k(k - 1)\left(\frac{c}{2}\right)^{k - 2} \end{align*}$

$\text{Since $c > 0$, $\left(\frac{c}{2}\right)^{k - 2} > 0$, the sign of $\cfrac{d^2y}{dx^2}$ matches the sign of $k(k - 1)$.} \\ \\ \text{Thus, if $0 < k < 1$, then $k(k - 1) < 0$ and so the stationary point at $\left(\frac{c}{2},\ 2\left(\frac{c}{2}\right)^k\right)$ is a maximum.} \\ \\ \text{And, if $k > 1$, then $k(k - 1) > 0$ and so the stationary point at $\left(\frac{c}{2},\ 2\left(\frac{c}{2}\right)^k\right)$ is a minimum.}$

$\text{Taking $c = a + b$ where $a > b > 0$ and considering the $0 < k < 1$ case, the maximum} \\ \\ \text{on $0 < x < a + b$ will be at $\left(\frac{a + b}{2},\ 2\left(\frac{a + b}{2}\right)^k\right)$. Since this is a local maximum:}$

$\begin{align*} y &< 2\left(\frac{a + b}{2}\right)^k \\ x^k + (a + b - x)^k &< 2\left(\frac{a + b}{2}\right)^k \qquad \text{as $c = a + b$} \\ \\ \text{Take $x = b$:} \quad b^k + (a + b - b)^k &< 2\left(\frac{a + b}{2}\right)^k \\ \frac{a^k + b^k}{2} &< \left(\frac{a + b}{2}\right)^k \qquad \text{as required} \end{align*}$

Luukas.2 · Dec 19, 2023

Aeonium said:
going to make a fool out of myself but split using log law –> ln(x^2e^(x^3)) - ln(sqrt(x^2+1)) –> diff each?

Yes, split with log laws... and more than you have suggested. But, imagine trying to do it without using log laws first, so you need to differentiate $\frac{x^2e^{x^3}}{\sqrt{x^2+1}}$ ...

Luukas.2 · Dec 19, 2023

The derivative is...

$\frac{2}{x} + 3x^2 + \frac{x}{x^2+1}$

Luukas.2 · Dec 20, 2023

@eternallyboreduser, does this solution make sense for the question you were looking at? The part that I have left out is very similar to the solution provided.

Luukas.2 said:
$\begin{align*} y &= x^k + (c - x)^k \\ \frac{dy}{dx} &= kx^{k - 1} + k(c - x)^{k - 1} \times -1 = k\left[x^{k - 1} - (c - x)^{k - 1}\right] \\ \\ \text{Put $\frac{dy}{dx} = 0$:} \quad k\left[x^{k - 1} - (c - x)^{k - 1}\right] &= 0 \\ \left(\frac{x}{c - x}\right)^{k - 1} &= 1 \qquad \text{as $k \neq 0$} \\ x &= \frac{c}{2} \\ \text{Put $x = \frac{c}{2}$:} \quad y &= \left(\frac{c}{2}\right)^k +\left(c - \frac{c}{2}\right)^k = 2\left(\frac{c}{2}\right)^k \\ \\ \text{Thus, the stationary point is at} \quad &\left(\frac{c}{2},\ 2\left(\frac{c}{2}\right)^k\right) \\ \\ \frac{d^2y}{dx^2} &= k(k - 1)x^{k - 2} - k(k - 1)(c - x)^{k - 2} \times -1 \\ &= k(k - 1)\left[x^{k - 2} + (c - x)^{k - 2}\right] \\ \\ \text{Put $x = \frac{c}{2}$:} \quad \frac{d^2y}{dx^2} &= k(k - 1)\left[\left(\frac{c}{2}\right)^{k - 2} + \left(c - \frac{c}{2}\right)^{k - 2}\right] = 2k(k - 1)\left(\frac{c}{2}\right)^{k - 2} \end{align*}$

$\text{Since $c > 0$, $\left(\frac{c}{2}\right)^{k - 2} > 0$, the sign of $\cfrac{d^2y}{dx^2}$ matches the sign of $k(k - 1)$.} \\ \\ \text{Thus, if $0 < k < 1$, then $k(k - 1) < 0$ and so the stationary point at $\left(\frac{c}{2},\ 2\left(\frac{c}{2}\right)^k\right)$ is a maximum.} \\ \\ \text{And, if $k > 1$, then $k(k - 1) > 0$ and so the stationary point at $\left(\frac{c}{2},\ 2\left(\frac{c}{2}\right)^k\right)$ is a minimum.}$

$\text{Taking $c = a + b$ where $a > b > 0$ and considering the $0 < k < 1$ case, the maximum} \\ \\ \text{on $0 < x < a + b$ will be at $\left(\frac{a + b}{2},\ 2\left(\frac{a + b}{2}\right)^k\right)$. Since this is a local maximum:}$

$\begin{align*} y &< 2\left(\frac{a + b}{2}\right)^k \\ x^k + (a + b - x)^k &< 2\left(\frac{a + b}{2}\right)^k \qquad \text{as $c = a + b$} \\ \\ \text{Take $x = b$:} \quad b^k + (a + b - b)^k &< 2\left(\frac{a + b}{2}\right)^k \\ \frac{a^k + b^k}{2} &< \left(\frac{a + b}{2}\right)^k \qquad \text{as required} \end{align*}$

difficult y11 diff qs? (2 Viewers)

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