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Question Number 4040 by Filup last updated on 27/Dec/15
f(x)=x^(sin^x (x)) g(x)=x Why is f ′(x)=0 on g(x)? Are all extrema (min, max, inflection) of f(x) on g(x)?
$${f}\left({x}\right)={x}^{\mathrm{sin}^{{x}} \left({x}\right)} \\ $$$${g}\left({x}\right)={x} \\ $$$$ \\ $$$$\mathrm{Why}\:\mathrm{is}\:{f}\:'\left({x}\right)=\mathrm{0}\:\mathrm{on}\:{g}\left({x}\right)? \\ $$$$ \\ $$$${Are}\:\boldsymbol{{all}}\:{extrema}\:\left(\mathrm{min},\:\mathrm{max},\:\mathrm{inflection}\right) \\ $$$$\mathrm{of}\:{f}\left({x}\right)\:\mathrm{on}\:{g}\left({x}\right)? \\ $$
Commented by Filup last updated on 27/Dec/15
Commented by Filup last updated on 27/Dec/15
So, when does: (df/dx)=g
$$\mathrm{So},\:\mathrm{when}\:\mathrm{does}: \\ $$$$\frac{{df}}{{dx}}={g} \\ $$
Commented by Yozzii last updated on 27/Dec/15
f(x)=x^(sin^x x) (∗) If f(x),x>0, taking ln on both sides of (∗) ⇒lnf(x)=(sin^x x)lnx (∗∗) Again, if lnf(x),(sin^x x)lnx>0, taking ln on both sidds of (∗∗) ⇒ln(lnf(x))=ln{(sin^x x)lnx} ln(lnf(x))=xlnsinx+ln(lnx) (∗∗∗) Implicitly differentiating (∗∗∗) w.r.t x gives (((d/dx)(lnf(x)))/(lnf(x)))=1×lnsinx+x×((cosx)/(sinx))+(((d/dx)(lnx))/(lnx)) (((1/(f(x)))f^( ′) (x))/(lnf(x)))=lnsinx+xcotx+((1/x)/(lnx)) ((f^( ′) (x))/(f(x)lnf(x)))=lnsinx+xcotx+(1/(xlnx)) f^( ′) (x)=f(x)lnf(x){lnsinx+xcotx+(1/(xlnx))} f^′ (x)=x^(sin^x x) lnx^(sin^x x) (1/(xlnx))(xlnx{lnsinx+xcotx}+1) f^′ (x)=((x^(sin^x x) (sin^x x)lnx(x(lnx){lnsinx+xcotx}+1))/(xlnx)) x,lnx>0 ∴f^′ (x)=x^(sin^x x−1) (sin^x x)(x(lnx){lnsinx+xcotx}+1) If f^′ (x)=0 ⇒sin^x x=0⇒x=nπ,x∈N f(nπ)=(nπ)^((sin{nπ})^(nπ) ) =(nπ)^0 =1 If g(x)=x touches f(x)=x^(sin^x x) it is necessary that ∃x∈R∣f^′ (x)=g^′ (x)=1.
$${f}\left({x}\right)={x}^{{sin}^{{x}} {x}} \:\:\:\:\:\:\left(\ast\right) \\ $$$${If}\:{f}\left({x}\right),{x}>\mathrm{0},\:{taking}\:{ln}\:{on}\:{both}\:{sides}\:{of}\:\left(\ast\right) \\ $$$$\Rightarrow{lnf}\left({x}\right)=\left({sin}^{{x}} {x}\right){lnx}\:\:\:\left(\ast\ast\right) \\ $$$${Again},\:{if}\:{lnf}\left({x}\right),\left({sin}^{{x}} {x}\right){lnx}>\mathrm{0},\: \\ $$$${taking}\:{ln}\:{on}\:{both}\:{sidds}\:{of}\:\left(\ast\ast\right) \\ $$$$\Rightarrow{ln}\left({lnf}\left({x}\right)\right)={ln}\left\{\left({sin}^{{x}} {x}\right){lnx}\right\} \\ $$$${ln}\left({lnf}\left({x}\right)\right)={xlnsinx}+{ln}\left({lnx}\right)\:\:\left(\ast\ast\ast\right) \\ $$$${Implicitly}\:{differentiating}\:\:\left(\ast\ast\ast\right)\:{w}.{r}.{t}\:{x} \\ $$$${gives}\: \\ $$$$\frac{\frac{{d}}{{dx}}\left({lnf}\left({x}\right)\right)}{{lnf}\left({x}\right)}=\mathrm{1}×{lnsinx}+{x}×\frac{{cosx}}{{sinx}}+\frac{\frac{{d}}{{dx}}\left({lnx}\right)}{{lnx}} \\ $$$$\frac{\frac{\mathrm{1}}{{f}\left({x}\right)}{f}^{\:'} \left({x}\right)}{{lnf}\left({x}\right)}={lnsinx}+{xcotx}+\frac{\mathrm{1}/{x}}{{lnx}} \\ $$$$\frac{{f}^{\:'} \left({x}\right)}{{f}\left({x}\right){lnf}\left({x}\right)}={lnsinx}+{xcotx}+\frac{\mathrm{1}}{{xlnx}} \\ $$$${f}^{\:'} \left({x}\right)={f}\left({x}\right){lnf}\left({x}\right)\left\{{lnsinx}+{xcotx}+\frac{\mathrm{1}}{{xlnx}}\right\} \\ $$$${f}^{'} \left({x}\right)={x}^{{sin}^{{x}} {x}} {lnx}^{{sin}^{{x}} {x}} \frac{\mathrm{1}}{{xlnx}}\left({xlnx}\left\{{lnsinx}+{xcotx}\right\}+\mathrm{1}\right) \\ $$$${f}^{'} \left({x}\right)=\frac{{x}^{{sin}^{{x}} {x}} \left({sin}^{{x}} {x}\right){lnx}\left({x}\left({lnx}\right)\left\{{lnsinx}+{xcotx}\right\}+\mathrm{1}\right)}{{xlnx}} \\ $$$${x},{lnx}>\mathrm{0} \\ $$$$\therefore{f}^{'} \left({x}\right)={x}^{{sin}^{{x}} {x}−\mathrm{1}} \left({sin}^{{x}} {x}\right)\left({x}\left({lnx}\right)\left\{{lnsinx}+{xcotx}\right\}+\mathrm{1}\right) \\ $$$${If}\:{f}^{'} \left({x}\right)=\mathrm{0}\: \\ $$$$\Rightarrow{sin}^{{x}} {x}=\mathrm{0}\Rightarrow{x}={n}\pi,{x}\in\mathbb{N} \\ $$$${f}\left({n}\pi\right)=\left({n}\pi\right)^{\left({sin}\left\{{n}\pi\right\}\right)^{{n}\pi} } =\left({n}\pi\right)^{\mathrm{0}} =\mathrm{1} \\ $$$$ \\ $$$${If}\:{g}\left({x}\right)={x}\:{touches}\:{f}\left({x}\right)={x}^{{sin}^{{x}} {x}} \\ $$$${it}\:{is}\:{necessary}\:{that}\:\exists{x}\in\mathbb{R}\mid{f}^{'} \left({x}\right)={g}^{'} \left({x}\right)=\mathrm{1}. \\ $$$$ \\ $$
Commented by Yozzii last updated on 27/Dec/15
Commented by Yozzii last updated on 27/Dec/15
The above graphs are g^′ (x)=1 and y=f′(x). We see that indeed ∃x∈R∣f^′ (x)=g^′ (x)=1 and that for such x they are in the neighbourhood of x=nπ at which f^′ (x)=0.
$${The}\:{above}\:{graphs}\:{are}\:{g}^{'} \left({x}\right)=\mathrm{1} \\ $$$${and}\:{y}={f}'\left({x}\right). \\ $$$${We}\:{see}\:{that}\:{indeed}\:\exists{x}\in\mathbb{R}\mid{f}^{'} \left({x}\right)={g}^{'} \left({x}\right)=\mathrm{1} \\ $$$${and}\:{that}\:{for}\:{such}\:{x}\:{they}\:{are}\:{in}\:{the}\: \\ $$$${neighbourhood}\:{of}\:{x}={n}\pi\:{at}\:{which}\: \\ $$$${f}^{'} \left({x}\right)=\mathrm{0}.\: \\ $$$$ \\ $$
Commented by Yozzii last updated on 27/Dec/15
No turning points of y=f(x) lie on y=g(x) as a consequence of the above graph showing g′(x)=f′(x)=1≠0 for g(x) touching f(x) near x=nπ.
$${No}\:{turning}\:{points}\:{of}\:{y}={f}\left({x}\right)\:{lie}\: \\ $$$${on}\:{y}={g}\left({x}\right)\:{as}\:{a}\:{consequence}\:{of}\: \\ $$$${the}\:{above}\:{graph}\:{showing}\:{g}'\left({x}\right)={f}'\left({x}\right)=\mathrm{1}\neq\mathrm{0} \\ $$$${for}\:{g}\left({x}\right)\:{touching}\:{f}\left({x}\right)\:{near}\:{x}={n}\pi. \\ $$
Commented by Filup last updated on 27/Dec/15
This is amazing and complicated! I don′t know implicite differentiation, but this is awesome!!!
$$\mathrm{This}\:\mathrm{is}\:\mathrm{amazing}\:\mathrm{and}\:\mathrm{complicated}! \\ $$$$\mathrm{I}\:\mathrm{don}'{t}\:\mathrm{know}\:\mathrm{implicite}\:\mathrm{differentiation}, \\ $$$$\mathrm{but}\:\mathrm{this}\:\mathrm{is}\:\mathrm{awesome}!!! \\ $$
Commented by Yozzii last updated on 27/Dec/15
This is why I′m glad we can use diagrams now. Graphing is important.
$${This}\:{is}\:{why}\:{I}'{m}\:{glad}\:{we}\:{can}\:{use} \\ $$$${diagrams}\:{now}.\:{Graphing}\:{is}\:{important}. \\ $$

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