Question

Find the first derivative of each of the following function and leave your answer in its simplest form:

Q1. y=2x^-5 + 3(sqrt x) - 8/7

dy/dx = -10x^-6 + 3/2 x ^(-1/2) ??
-------------------------------------
Q2)). r=((t³-2)/(4t⁵+19))
U=(t^3)-2 v= 4t^5 +19
du/dx=3t^2 dv/dx=20t^4

(v(du/dx)- u(dv/dx))/v^2

(4t^5 +19)(3t^2)-((t^3)-2 )(20t^4)
------------------------------
( 4t^5 +19)^2

12t^7+57t^2 - 20t^7 + 40t^4
----------------------------
( 4t^5 +19)^2

-8t^7+40t^4-57t^2
------------------
( 4t^5 +19)^2

t^2(-8t^5+40t^2-57)
-------------------
( 4t^5 +19)^2

Answer 1

is my answers correct?

Answer 2

@hartnn @UnkleRhaukus @jennychan12 @AccessDenied

Answer 3

the first one is right!

Answer 4

how about Q2?

Answer 5

i can't see any mistake,

Answer 6

was that the reason i been ignored for so many hours ><

Answer 7

maybe because your question isn't in \(\LaTeX\)

Answer 8

anyway thank you uncle

Answer 9

i think you can simply the second one a bit

Answer 10

Q2)\[r=\frac{t^3-2}{4t^5+19}\]
\[u=t^3-2\qquad v= 4t^5 +19\]\[\frac{du}{dx}=3t^2\qquad\frac{dv}{dx}=20t^4\]
\[\begin{align*}
\frac{dr}{dx}&=\frac{v(du/dx)- u(dv/dx)}{v^2}\\
&=\frac{(4t^5 +19)(3t^2)-(t^3-2 )(20t^4)}{( 4t^5 +19)^2}\\
&=\frac{12t^7+57t^2 - 20t^7 + 40t^4}{( 4t^5 +19)^2}\\
&=\frac{-8t^7+40t^4-57t^2}{( 4t^5 +19)^2}\\
&=\frac{t^2(-8t^5+40t^2-57)}{( 4t^5 +19)^2}\end{align*}\]

Answer 11

ya!! that why my final answer is
t^2(-8t^5+40t^2-57)
-------------------
( 4t^5 +19)^2

Answer 12

actually you won't be able to simplify,
but now i can see one tiny minus sign error

Answer 13

can you spot it?

Answer 14

on the 57

Answer 15

u refering to +57?

Answer 16

yeah