The function $f$ is defined by $f(t)= \frac{4e^{kt} - 1}{4e^{kt} +1}$, where $k$ is a positive constant.
(a) Show that $f'(t)>0$.
(b ) Show that $k(1- [f(t)]^2) = 2f'(t)$ and, hence, show that $f''(t)<0$.
$f(t)= \frac{4e^{kt} - 1}{4e^{kt} +1}$, where $k$ is a positive constant.
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ordinary-differential-equations
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0can you use $$\LaTeX$$ please? – 2017-01-14
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0What formulas for computing derivatives have you learned? Do you know the derivative of $e^{kt}$? Try writing the formula for $f'(t)$, it looks like you will need it for part (b) and it may be the easiest way to do part (a) as well. – 2017-01-15
2 Answers
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differentiating $$f(t)$$ with respect to $$t$$ we get after simplification $$f'(t)=8\,{\frac {k{{\rm e}^{kt}}}{ \left( 4\,{{\rm e}^{kt}}+1 \right) ^{2}}}$$ since $k>0$ and the exponential function is poesitive we get $$f'(t)>0$$ for all $t$ simplifying the equation $$k(1-f(t)^2)$$ we obtain $$16\,{\frac {k{{\rm e}^{kt}}}{ \left( 4\,{{\rm e}^{kt}}+1 \right) ^{2}} } $$ and this is $$2f'(t)$$ and we get for the second derivative as $$f''(t)=-8\,{\frac {{k}^{2}{{\rm e}^{kt}} \left( 4\,{{\rm e}^{kt}}-1 \right) }{ \left( 4\,{{\rm e}^{kt}}+1 \right) ^{3}}} $$
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(a)$$f'(t)=\frac{(4e^{kt}-1)' \cdot (4e^{kt}+1)-(4e^{kt}-1)\cdot(4e^{kt}+1)'}{(4e^{kt}+1)^2}$$
$$=\frac{4ke^{kt}(4e^{kt}+1)-4ke^{kt}(4e^{kt}-1)}{(4e^{kt}+1)^2}$$ $$=\frac{16ke^{2kt}+4ke^{kt}-16ke^{2kt}+4e^{kt}}{(4e^{kt}+1)^2}$$ $$=\frac{8ke^{kt}}{(4e^{kt}+1)^2}>0$$ since $k$ is a positive constant.
(b)$$k(1-f(t)^2)=k\Bigl(1-\frac{(4ke^{kt}-1)^2}{(4ke^{kt}+1)^2}\Bigr)$$ $$=k\Bigl(\frac{(4e^{kt}+1)^2-(4e^{kt}-1)^2}{(4e^{kt}+1)^2}\Bigr)$$ $$=k\Bigl(\frac{(4e^{kt}+1+4^{kt}-1)\cdot(4e^{kt}+1-4e^{kt}+1)}{(4e^{kt}+1)^2}\Bigr)$$ $$=k\cdot\frac{8e^{kt} \cdot 2}{(4e^{kt}+1)^2}$$ $$=\frac{16ke^{kt}}{(4e^{kt}+1)^2}=2 \cdot f'(t)$$
Also,
$$f''(t)=\frac{(8ke^{kt})' \cdot (4e^{kt}+1)^2 - (8ke^{kt}) \cdot [(4e^{kt}+1)^2]'}{(4e^{kt}+1)^4}$$ $$=\frac{8k^2e^{kt}(4e^{kt}+1)^2-(8ke^{kt})\cdot\bigl(2\cdot(4e^{kt}+1)\cdot(4e^{kt}+1)'\bigr)}{(4e^{kt}+1)^4}$$ $$=\frac{8k^2e^{kt}\cdot(4e^{kt}+1)^2-16ke^{kt}\cdot (4e^{kt}+1)(4ke^{kt})}{(4e^{kt}+1)^4}$$ $$=\frac{8k^2e^{kt}(4ke^{kt}+1)-16ke^{kt} \cdot 4ke^{kt}}{(4e^{kt}+1)^3}$$ $$=\frac{8k^2e^{kt}[4e^{kt}+1-8e^{kt}]}{(4e^{kt}+1)^3}$$ $$\frac{8k^2e^{kt}(-4e^{kt}+1)}{(4e^{kt}+1)^3}<0$$ since $-4e^{kt}+1<0$.