c4 easy 2 doubts

[zabady]

i have this doubt in the proof of a^x=INa

i kno that y=a^x
log both sides =

INy=xINa

differentiation of iny is easy but when it comes to differentian of xINa i do it by product rule which is X(1/a)+INa

in the book the ignored the x(1/a) thats where i dnt understand i attached the explanation of the book

my second doubt is attached too

[astarmathsandphysics]

Xlna is not a product. Lna is a constant

[zabady]

aha so a constant we would take it as a zero when we differentiate it i have still 1 doubt which is the second picture

[T.Q]

FOR the first doubt , u dont have to know how to derive it
just memories it  like a^x =   dy/dx=a^xlna


and for the second doubt, u differentiate the whole equ.

2x+8y(dy/dx)-6-16(dy/dx)=0

make dy/dx subject of the formula

dy/dx=(6-2x)/(8y-16)

dy/dx=0

so , x=3

now substitute the x value in the original equ and solve

y= 1  or 3


so the coordinates are (3,1) , (3,3)

[zabady]

ooooh dam i kno that one i gave the wrong question look at the one i attached here i tried solving it same way as the one u just solved but it gives t=1 and in question they say t is not equal to one so how to solve it 

[T.Q]

ooooh dam i kno that one i gave the wrong question look at the one i attached here i tried solving it same way as the one u just solved but it gives t=1 and in question they say t is not equal to one so how to solve it 

okay


dx/dt= 1/(1-t)^2    &    dy/dt= t(2-t)/(1-t)^2

so divide dy/dt by dx/dt  to give u dy/dx

so , dy/dx= t(2-t)

t= 0 or 2


so the coordinates are

(0,0)  ,   (-2,-4)

[zabady]

thaaankz so much man sry for wastin ur time it waz easy thankz again mannn                 

[T.Q]

no problem   

im here to help 

[Amr Fouad]

thanks T.Q

+rep

[T.Q]

thanx ~ A.F ~