Question

f(x)=e^x sinx => f '(0)=?

Answer 1

f'(x)=e^xsinx+e^xcosx
thus f'(0)=e^0sin0+e^0cos0=1

Answer 2

f'(U*V) =U f'(V) + V f'(U)

Answer 3

The first step is to find the function for the derivative, which can be found using the product rule:
\( (h(x) * g(x))' = h'(x) g(x) + h(x) g'(x) \)
\( (e^x \sin x)' = e^x \sin x + e^x \cos x \)
Then, the notation \(f'(0)\) means to plug in \(x=0\) into the function for the derivative.

Answer 4

if, \( f(x) = e^x g(x) \)
then \( f'(x) = e^x(g(x) + g'(x)) \)

Answer 5

\[\int\limits \sqrt{x-1} dx\]

Answer 6

try using trigonometric substitution ... x = sec^2 u
or try substituting x - 1 = t^2, dx = 2tdt

Answer 7

but whay is dx=2tdt

Answer 8

Couldn't you just use a u-substitution with \(u = x - 1\) ?
Then you can reduce it to an integral of a variable to a rational power...

Answer 9

yes,
dx=2tdt
=integral 2t^2dt=2/3t^3
=\[2/3(\sqrt{x-1})^3\]

Answer 10

the answer here is f'(x)=1 , isn't it?

Answer 11

oh, i didn't see the next qsn.. :P

Answer 12

\[\int\limits_{}^{}2t^2dt=2/3t^3=2/3(\sqrt{x-1})^3\]

Answer 13

\[\LARGE \begin{array}{l}f\left( x \right) = {e^x} \cdot \sin x\\f'\left( x \right) = {\left( {{e^x} \cdot \sin x} \right)^'}\\f'\left( x \right) = {\left( {{e^x}} \right)^'} \cdot \sin x + {e^x} \cdot {\left( {\sin x} \right)^'}\\f'\left( x \right) = {e^x} \cdot \sin x + {e^x} \cdot \cos x\\f'\left( x \right) = {e^x}\left( {\sin x + \cos x} \right)\\\\f'\left( 0 \right) = {e^0}\left( {\sin 0 + \cos 0} \right)\\f'\left( 0 \right) = 1 \cdot \left( {0 + 1} \right)\\f'\left( 0 \right) = 1\end{array}\]

Detyra e dyte shkon...
\[\LARGE \int \sqrt{x-1}\;\;dx\]
zevendesojme \(x-1\) me \(t^2\) pastaj derivojme dy anet e barazimit...
\[\left| \LARGE {\begin{array}{*{20}{c}}{x - 1 = {t^2}}\\{{{\left( {x - 1} \right)}^'} = {{\left( {{t^2}} \right)}^'}}\\{1 \cdot dx = 2t \cdot dt}\end{array}} \right|\]
e tasht ne vend te x-1 e shtim \(t^2\) dhe ne vend te \(dx\) kemi \(2tdt\) :
\[\LARGE \int\sqrt{t^2}\;\;2tdt\]
\[\LARGE \int t\;\;2tdt\]
\[\LARGE 2\int t^2 dt\]
pjesa tjeter eshte kallaj.... @Labinot