Question

calc

Answer 1

this is the same concept we just did :)

Answer 2

use chain rule :)

Answer 3

so u know chain rule?

Answer 4

\[\frac{ d }{ dx }[f[g(x)]]=f'(x)*\frac{ d }{ dx }g(x)=f'(x).g'(x)\]

Answer 5

\[\frac{dy}{dx}=\frac{dy}{d[x^2+e^{3x}]}*\frac{d[x^2+e^{3x}]}{dx}\]

Answer 6

ok
\[f(x)=\ln(x^2+e^{3x})\\f'(x)=\frac{ d }{ dx }\ln(x^2+e^{3x})=\frac{ 1 }{ x^2+e^{3x} }*\frac{ d }{ dx }(x^2+e^{3x})\]
ok so far?

Answer 7

\[\frac{ d }{ dx }\ln(x)=1/x\]
used this

Answer 8

now just complete the rest
\[f'(x)=\frac{ 1 }{ x^2+e^{3x} }[\frac{ d }{ dx }(x^2)+\frac{ d }{ dx }(e^{3x})]=\frac{ 1 }{ x^2+e^{3x} }[2x+3e^{3x}]\]

Answer 9

\[\frac{ d }{ dx }e^{ax}=ae^{ax}\]
used this

Answer 10

got it?

Answer 11

no its complete lol :)

Answer 12

yw!!