Title: IMA1 (viikko 6) muut

Find the derivatives of the following functions: Määritä seuraavien funktioiden derivaatat:

(a) \[f(x)=[value]\]

(b) \[g(x)=[value]\]

(c) \[h(x)=[value]\]

(a)\(\quad f'\left(x\right)=\)

(b)\(\quad g'\left(x\right)=\)

(c)\(\quad h'\left(x\right)=\)
Feedback: Model solution: Mallivastaus: (a) Chain rule: Ketjusääntö: \[(f\circ g)'(x)=f'(g(x))g'(x): \] \[f(x)=([value]-[value])^{[value]} \] \[ f'(x) =[value]([value]-[value])^{[value]}([value]x^{[value]}-[value]x^{[value]}) \] \[=[value]([value])^{[value]}(1-[value])^{[value]}x^{[value]}([value]-[value]x^{[value]}) \] \[=[value](1-[value])^{[value]}x^{[value]}([value]-[value]x^{[value]}) \] \[=[value]x^{[value]}(1-[value])^{[value]}([value]-[value]x^{[value]}).\] (b) Quotient rule: Osamäärän derivaatta: \[\Big( \frac{f}{g}\Big) '(x)=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}: \] \[g(x)=\frac{x^3+1}{\sqrt{1+x^2}}=\frac{x^3+1}{(1+x^2)^{1/2}} \] \[g'(x)=\frac{3x^2\sqrt{1+x^2}-(x^3+1)\frac{1}{2}(1+x^2)^{-1/2} 2x}{[(1+x^2)^{1/2}]^2 } \] \[=\frac{3x^2\frac{1+x^2}{\sqrt{1+x^2}}-\frac{x(x^3+1)}{\sqrt{1+x^2}}}{1+x^2} \] \[=\frac{3x^2+3x^4-x^4-x}{(1+x^2)^{3/2}}=\frac{2x^4+3x^2-x}{(1+x^2)^{3/2}}. \] (c) Product rule: Tulon derivaatta: \[(fg)'(x)=f'(x)g(x)+f(x)g'(x). \] \[h(x)=\cos([value]x)\sin([value]x) \] \[h'(x)=-\sin([value]x)\cdot [value]\cdot \sin([value]x)+\cos([value]x)\cdot \cos([value]x)\cdot [value] \] \[=-[value]\cdot \sin([value]x)\cdot \sin([value]x)+[value]\cdot \cos([value]x)\cdot \cos([value]x). \]

Laske seuraavat integraalit

(a)\[\int_{[value]}^{[value]}([value]x-[value]\sqrt{x}+[value]) \, \textrm{d}x \] (b)\[\int_{0}^{[value]} \cos(\frac{x}{2}) \, \textrm{d}x \] (c)\[\int_{[value]}^{[value]} [value] \, \textrm{d}x \]

Anna vastaukset tarkkoina arvoina! Neliöjuuri a merkitään sqrt(a), esim. sqrt(2). Luonnollisen logaritmin kantaluku e merkitään e, esim. 2*e.

(a) | \(\int_{[value]}^{[value]}([value]x-[value]\sqrt{x}+[value]) \, \textrm{d}x=~\)
(b) | \(\int_{0}^{[value]} \cos(\frac{x}{2}) \, \textrm{d}x=~\)
(c) | \(\int_{[value]}^{[value]} [value] \, \textrm{d}x=~\)
Feedback: Mallivastaus: (a) \begin{eqnarray*} \int_{[value]}^{[value]}([value]x-[value]\sqrt{x}+[value]) \, \textrm{d}x & = & \bigg/_{\!\!\!\![value]}^{\,[value]}([value]x^2-[value]\cdot \frac{2}{3}x^{3/2}+[value]x) \\ & = & [value]\cdot [value]^2-[value]\cdot [value]^{3/2}+[value]\cdot [value]-[value]\cdot [value]^2+[value]\cdot [value]^{3/2}-[value]\cdot [value] \\ & = & [value] \end{eqnarray*} (b) \[\int_{0}^{[value]} \cos(\frac{x}{2}) \, \textrm{d}x=\bigg/_{\!\!\!\!0}^{\,[value]}2\sin(\frac{x}{2})=2\sin([value])-2\sin(0)=2\sin([value])=[value]. \] (c) \[\int_{[value]}^{[value]} [value] \, \textrm{d}x=\bigg/_{\!\!\!\![value]}^{\,[value]} [value]=[value]. \]

Laske integraalit \[\text{(a) } \int_{0}^[value] [value] \, \mathrm{d}x\] \[\text{(b) } \int_{0}^[value] [value] \, \mathrm{d}x\] (Vihje: Mikä on \(D(\ln(\cos(x))\)?)

Anna vastauksesi tarkkoina arvoina. Luonnollinen logaritmi a:sta merkitään ln(a), esim. ln(2). Luonnollisen logaritmin kantaluku merkitään e, esim. 2*e. Neliöjuuri a:sta merkitään sqrt(a), esim. sqrt(2).

Vastaukset:

(a) |  |
(b) |  |
Feedback: Mallivastaus: (a) \begin{eqnarray*} \int_{0}^{[value]} |[value]-x| \, \mathrm{d}x & = & \int_{0}^{[value]} ([value]-x) \, \mathrm{d}x+\int_{[value]}^{[value]} (x-[value]) \, \mathrm{d}x \\ & = & \bigg/_{\!\!\!\!0}^{\,[value]} ([value]x-\frac{1}{2}x^2)+\bigg/_{\!\!\!\![value]}^{\,[value]} (\frac{1}{2}x^2-[value]x) \\ & = & [value]\cdot [value]-\frac{1}{2}\cdot [value]^2+\frac{1}{2}\cdot [value]^2-[value]\cdot [value]-\frac{1}{2}\cdot [value]^2+[value]^2 \\ & = & [value]. \end{eqnarray*} (b) \[D\left(\ln\left(\cos(x)\right)\right)=\frac{1}{\cos(x)}(-\sin(x))=\frac{-\sin(x)}{\cos(x)}=-\tan(x).\] \[\int_{0}^{[value]} \tan(x) \, \mathrm{d}x = \bigg/_{\!\!\!\!0}^{\,[value]} -\ln\left(\cos(x)\right)=-\ln\left(\cos\left([value]\right)\right)-\left[-\ln\left(\cos(0)\right)\right] = [value].\]

Määritä seuraavat integraalit. Anna vastauksesi tarkkana arvona. Neliöjuuri merkitään \(\sqrt{x} = \mathtt{sqrt(x)}\) ja \(\pi = \mathtt{pi}\).

(a) | \(\displaystyle \int_{[value]}^{[value]} \left( [value]\right) \, \mathrm{d}x =~\) |
(b) | \(\displaystyle \int_{[value]}^{[value]} [value] \, \mathrm{d}x =~\) |
Feedback: Mallivastaus: (a) \[\begin{array}{ll} & \displaystyle{ \int_[value]^[value] \left([value]\right) \mathrm{d}x} \\ = & \displaystyle{ \Bigg/_{\!\!\!\![value]}^{\,[value]} \left( [value]\right)} \\ = & \displaystyle{ [value]\cdot[value]^2- +[value]\cdot[value]^{\frac{3}{2}}- +[value]\cdot[value] - \left([value]\cdot[value]^2- +[value]\cdot[value]^{\frac{3}{2}}- +[value]\cdot[value] \right)} \\ = &\displaystyle{ [value]} \end{array}\] (b) \[\begin{array}{rll} \displaystyle\int_[value]^[value] [value] \mathrm{d}x & = & \displaystyle \Bigg/_{\!\!\!\![value]}^{\,[value]} \left( [value] \right)\\ & = &\displaystyle -[value]\cos\left([value]\right) - \left(-[value] \cos\left([value]\right)\right) \\ & = & \displaystyle [value] + - [value] \sqrt{3} \sqrt{2} \frac{3}{\sqrt{2}} \end{array}\]

Laske määrätyt integraalit.

(a) \(\displaystyle \int\limits_[value]^[value] \left( [value] \right) \mathrm{d}x = \)

(b) \( \displaystyle \int\limits_[value]^[value] [value] x\sqrt{x} \, \mathrm{d}x = \)

(c) \( \displaystyle \int\limits_[value]^[value] [value] \, \mathrm{d}x = \)
Feedback: Mallivastaus: (a) \( \displaystyle \int\limits_[value]^[value] [value] \, \mathrm{d}x = {\bigg/_{\mspace{-15mu}[value]}^{\,[value]}} \left( [value] \right) = [value] \) (b) \( \displaystyle \int\limits_[value]^[value] [value] x\sqrt{x} \, \mathrm{d}x = {\bigg/_{\mspace{-15mu}[value]}^{\,[value]}} [value] x^{\frac{5}{2}}= [value] \) (c) \( \displaystyle \int\limits_[value]^[value] [value] \, \mathrm{d}x = {\bigg/_{\mspace{-15mu}[value]}^{\,[value]}} [value] \tan x = [value] \)

Laske määrätyt integraalit
\( \textbf{(a)} \; \displaystyle \int\limits_[value]^[value] \left( [value] \right) \mathrm{d}x = \)
\( \textbf{(b)} \; \displaystyle \int\limits_[value]^[value] [value] x\sqrt{x} \, \mathrm{d}x = \)
\( \textbf{(c)} \; \displaystyle \int\limits_[value]^[value] [value] \, \mathrm{d}x = \)
Feedback: Mallivastaus: \( \textbf{(a)} \) \( \displaystyle \int\limits_[value]^[value] [value] \, \mathrm{d}x = {\bigg/_{\mspace{-15mu}[value]}^{\,[value]}} \left( [value] \right) = [value] \) \( \textbf{(b)} \) \( \displaystyle \int\limits_[value]^[value] [value] x\sqrt{x} \, \mathrm{d}x = {\bigg/_{\mspace{-15mu}[value]}^{\,[value]}} [value] x^{\frac{5}{2}}= [value] \) \( \textbf{(c)} \) \( \displaystyle \int\limits_[value]^[value] [value] \, \mathrm{d}x = {\bigg/_{\mspace{-15mu}[value]}^{\,[value]}} [value] \tan x = [value] \)