Question

Calc (Derivatives) am I doing these problems correctly? (teacher doesn't want us to simplify) (question in comments)

Answer 1

\[y=\sqrt{4x^2 + x -3}\]
\[y'(x) = 4x^2 +x -3) ^(1/2)\]
then, using the power rule
\[= (1/2)(4x^2 + x - 3)^(-1/2) (8x+1)\]

Answer 2

\(= (1/2)(4x^2 + x - 3)^{-1/2} (8x+1)\)

Answer 3

like that ?

Answer 4

\[y=(7x-92)^\frac{ 1 }{ 4 } \]
to \[\frac{ 1 }{ 4 }(7x-92)^\frac{ 3 }{ 4 }(7)\]

Answer 5

yeah, sorry

Answer 6

first one is correct

Answer 7

for second one, check the sign on exponent...

Answer 8

Ok, cool

Answer 9

\(\frac{ 1 }{ 4 }(7x-92)^\frac{ \color{red}{-}3 }{ 4 }(7) \)

Answer 10

oh, yeah. I forgot to make it negative :)

Answer 11

Can I then multiply (1/4) and 7?

Answer 12

absolutely !

Answer 13

cool! thanks :)

Answer 14

Tell me, how much harder does calculus get?

Answer 15

oh, so, on the second Problem I have (-7/4)(7x-92)^(-3/4) when I move the negative exponent down, does the problem read \[f'(x)=\frac{ -7 }{ 4 }/[(7x-92)^\frac{ 3 }{ 4 }]\] or \[\frac{ -7 }{ 4(7x-92)^\frac{ 3 }{ 4 } }\] ?

Answer 16

Sorry for formatting, they are both supposed to look like the second example

Answer 17

hey no

Answer 18

\(\large y=(7x-92)^\frac{ 1 }{ 4 } \)

\(\large y'=\frac{1}{4}(7x-92)^{\frac{ 1 }{ 4 }-1} (7) \)

\(\large y'=\frac{7}{4}(7x-92)^{\frac{ -3}{ 4 }} \)

\(\large y'=\frac{7}{4(7x-92)^{\frac{ 3}{ 4 }}} \)

Answer 19

there will not be any negative signs...
negative sign is in exponent only when we first differentiate..

Answer 20

Thank you! I don't know where that negative came from lol. Thanks again!