Question

Calculus help!! MEdal!!

Answer 1

and where are you stuck?

Answer 2

i don't know where to begin to solve this problem

Answer 3

\(\text{derivative of integration} \\
\Large \dfrac{d}{dx}\int \limits_{f(x)}^{g(x)}h(x)dx = h(g(x))g'(x)-h(f(x))f'(x)\)

here, f(x) is a constant, so 2nd term will not be present

Answer 4

what 2nd term?

Answer 5

\(\\
\Large \dfrac{d}{dx}\int \limits_a^{g(x)}h(x)dx = h(g(x))g'(x)\)

Answer 6

oh you mean f(x)

Answer 7

so i should just set up my problem like this and solve?

Answer 8

yes

Answer 9

if you compare,
a = 2,
g(x) = x^2
h(x) = \((e^x)^3\)

Answer 10

oh ok i'll give it a try

Answer 11

let me know what u get..

Answer 12

wait what is g'(x)?

Answer 13

g(x) = x^2

g'(x) is the derivative of g(x) = ...

Answer 14

oh okay just making sure

Answer 15

h(g(x)) g'(x)
e(x^2)^3 x^2
x^6e 2x
2x^7e

Answer 16

its not
\(e\times x^2\)

its
\(e^{x^2}\)

Answer 17

ohhh right

Answer 18

\(\implies (e^{x^2})^3 \times (x^2)' \\ \Large = e^{x^{2\times 3}}\times 2x\)
makes sense?

Answer 19

yes

Answer 20

let me try to do that

Answer 21

i got 2xe^(3x)^2

Answer 22

laws of exponent
\(\Large (A^B)^C = A^{BC}\)

Answer 23

oh wait no

Answer 24

12e^x^6 x^6 + 2e^x^6

Answer 25

\(\implies (e^{x^2})^3 \times (x^2)' \\ \Large = e^{x^{2\times 3}}\times 2x\)

is just
\(\Large 2x{e^x}^6\)