Question

Need to find f'(0) of this function

Answer 1

\[\frac{ e^x g(x) }{ h(x)+2g(x) }\]

Answer 2

g(0) is 2, g'(0) is 5, h(0) is 1, h'(0) is -1

Answer 3

do u know the quotient rule?

Answer 4

oh im dumb plug it in of g(x) with x as 0, and then derive the resulting thing, (its very easy)

Answer 5

\[\frac{ e^x g(x) }{ h(x)+2g(x) }\]
\[ e^x~ g~ ({ h+2g })^{-1}\]
\[ e^x~ g~ ({ h+2g })^{-1}\\
+e^x~ g'~ ({ h+2g })^{-1}\\
+e^x~ g~ ({ h+2g })'^{-1}\]
\[({ h+2g })'^{-1}=-({ h+2g })^{-2}~( h'+2g')\]

Answer 6

quotient rule is overrated, product rule is much simpler to keep track of

Answer 7

x = 0
g=2, g'=5
h=1, h'=-1

plug and play

Answer 8

cant u differentiate
e^x * 2 / (1+2*2)
??
@amistre64

Answer 9

and then u get 2/5 e^x as ur derivative

Answer 10

well, the functions are not constant in their definitions, but lets see what out final result is

Answer 11

yah but its only the derivative at 0 so?

Answer 12

\[ 2/~ ({ 1+2*2 })\\
+ 5/~ ({ 1+2*2 })\\
- 2~ (-1+2*5)/({ 1+2*2 })'^{2}\]

\[ 2/5+ 5/5
- 18/25\]

Answer 13

so, 7/5 - 18/25

(35- 18)/25

17/25 not= 2/5

Answer 14

so no, this is not the same as the derivative of 2/5 e^x

Answer 15

in otherwords, the function at x=0 is 2/5, but the slope of the line at x=0 is 17/25 assuming of course i didnt make any errors