Diffrenciate wt respect to x
y=1/sqrtX

u cant differentiate wrt an parameter if it is not in the equtn lik i cant see any 'x' in ur equation.

how possible is it to differentiate just ordinary square root. i guess you owe me an explanation on that if possible

Typing error dudes bt dat has bin corrected

it can also be written as
y=1/x^1/2
therefore y=x^-1/2
dy/dx=-1/2{x^-1/2-1}
=-1/2{x^3/2}
= -1/{2x^3/2}


i guess its not more than this is it?

Very true Uzzy, u r absolutely rite. U fit teach me integration

thats my area of specialization so u are so welcome to mi class!!!

dats was really cheap, i need a better one to crack

I am not trying to be Mr Too-know , none of you can solve these questions. I double dare you guys.
Thees are it:
find the value for X in:
1. 3^x + 4^x = 5^x

2. Prove that aⁿ = bⁿ + cⁿ has no nonzero integer solutions for n>2.

Rule: do not consult a professional mathematician, like a leaturer, only consult you best math textbooks

Also play with my little friends:

1000kgs equals ton thn3000kgs equals?

What comes after Ten Crores?

What is the lowest number with ten prime factors?

How much is five eighth's of a cup?

Bet mi i have bin longing to ask this question since i have been in secondary school but i guess i would try solving it and if i cant one of u shld. Now u were faster than mi there and annoyingly u know the answer!!!

The most annoying thing is that i have always know the answer to be 2 but have really tried solving it out. *sobbing*

@Pongwa
Beta get this one right!!

logarithmically(if dat exist) it is roughly 2, 1.54 to be precise(d closest i got). Tk log of both sides, factorise nd bring out x(l.h.s),divide thru by l.h.s,u shld v one left on d l.h.s, factorise for x on d r.h.s, gt a value for x shld be 0.6 or 1.54 dependn on hw u go abt it, d 2nd questn wil be solved wt a knowledge of d 1st(looks louely tho). Hope i tried

you can diff. it from using the first principle. the answer to the math is 1/2rtx rem rt means root

if y=1/rtx,
rem the formula for diff. i.e dy/dx=nx^n-1
therefore,
dy/dx=1/2x^1/2-1

=1/2x^-1/2

=1/2of 1/rtx

=1/2rtx QED
If u have any question pls feel free to reach d prof

pongwa:

logarithmically(if dat exist) it is roughly 2, 1.54 to be precise(d closest i got). Tk log of both sides, factorise nd bring out x(l.h.s),divide thru by l.h.s,u shld v one left on d l.h.s, factorise for x on d r.h.s, gt a value for x shld be 0.6 or 1.54 dependn on hw u go abt it,

@pongwa   taking log. of both sides in the equation, the answer is still gonna arrive at Zero. i will really appreciate it if the poster of this question could give the solution with workings since no efiko online has been able to solve for x in the eqn. looking forward to hear from you POSTER.

no ur wrong, dividing both sides by l.h.s will leave 1 on one side. check ur math

My people this is a very simple math. you don't have to solve for x. This is differentiation. You solve for dy/dx. Since y has been given,  to differentiate wrt x, it means dy/dx. Don't confuse youself.

when y=1/sqrtx,
then dy/dx= x^-1/2
applying the formula,
= -1/2x^-3/2
Therefore, the answer to tthe math is= -1/2cuberootx.

i still disagree with you pongwa. why dont you work it out in calculation here that would be of help to me and others. cos from what am seeing from this equation the answer is zero.

pls what does that ^ stand for?

,,

napolie
xlog3 + xlog4 eq xlog5
x(log3+l0g4) eq xlog5

1 eq xlog5/ x(log3 +log4)
1 eq x( log5/log3+log4)




finish up and tell me wat u got

cos(5x-28) eq sin(3x-50) since both are acute remember cos(90-x) eq sinx
therefore
sin(90-(5x-28)) eq sin (3x-50) sin vanishes
118-5x eq 3x-50
168 eq 8x
x eq 21 degrees, since it is acute thats why i solved it dt way

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1 substitution depends on which side ur workin on, if u decide to substitute the right sides then it will become cos and not sin like i did, u will still get the answer u shld try it on ur own.
2 there is a general formular for that i think u shld consult a text ,and also u can use ur knowledge of elememtary trigonometry signs(ACTS)

.

i dont understand the question

p.o.c

japro, i'm sorry but wt& is pile? do you mean pi?

.

{2sin2xdx is not difficult to integrate. Since the pi/2 & pi/8 are constants you can choose to multiply those in later.

-cos2x + arbitrary indefinite integral constant C

[2sin2xdx.

let u = 2x
dx = du/2

[2sinudu/2 = [sinudu = -cosu + c

sub

-cos2x + C

smh when does school end for u sef?

meh in like 3 weeks. i have a test 2marrow sef that's why i'm still up studyinf