Question

Anyone good at Calculus??

Answer 1

kind off

Answer 2

sometimes!

Answer 3

5 is correct as its a special integral f'(x) / f(x)

Answer 4

hmm that doesn't look like calculus =P

Answer 5

well, at least the first two

Answer 6

4 is b - similar integral as in 5

Answer 7

is number 2 ln (e^4) ?

Answer 8

if so it = 4 ln e which is what?

Answer 9

ln e = ?

Answer 10

recall one rule of logs:- loga a = 1

Answer 11

im not sure about 1

Answer 12

\(7ln(x)-8ln(x^2+6)\implies ln(x^7)-ln[(x^2+6)^8]\implies ln\left[ \cfrac{x^7}{(x^2+6)^8} \right]\)

Answer 13

I'.ve shown you how to do 2

Answer 14

4 ln e
and ln e simplifies to what valuye?

Answer 15

oh its A not 4

Answer 16

or rather \(\large {
log_{\color{red}{ a}}{\color{red}{ a}}^x\implies x\qquad thus
\\ \quad \\
ln(e^A)\implies log_{\color{red}{ e}}{\color{red}{ e}}^A\implies A
}\)

Answer 17

ln e = 1

Answer 18

\(7ln(x)-8ln(x^2+6)\implies ln(x^7)-ln[(x^2+6)^8]\implies ln\left[ \cfrac{x^7}{(x^2+6)^8} \right]\)

Answer 19

yes I got that too

Answer 20

for number 3 you need to use the Chain /Quotent rule I believe.

Answer 21

hold the mayo

Answer 22

hmm i'll have to do this on paper

Answer 23

any thoughts?

Answer 24

well I could post what I have so far, one sec

Answer 25

hmm i#m sure i could do this but im too tired)

Answer 26

Looking at your math for #1 it doesnt have a correct answer?!?!?!?!?!?!

Answer 27

\(\bf \cfrac{dy}{dx}=\left( \cfrac{1}{x\sqrt{x^2+15}} \right)
\left( \sqrt{x^2+15}+x\left( \frac{1}{\cancel{2}}(x^2+15)^{-\frac{1}{2}}\cdot \cancel{2} x \right) \right)
\\ \quad \\

\cfrac{dy}{dx}=\left( \cfrac{1}{x\sqrt{x^2+15}} \right)\left( \sqrt{x^2+15}+\cfrac{x^2}{\sqrt{x^2+15}} \right)\)

thus far