Differentiate exponential

[wind]

Hi can someone check over my work? thanks


Differentiate

\(\displaystyle \L\ y= x^{3} e^{-x}\)

f(x)= \(\displaystyle \L\ x^{3}\)
f'(x)= \(\displaystyle \L\ 3x^{2}\)

g(x)=\(\displaystyle \L\ e^{-x}\)
g'(x)=\(\displaystyle \L\ e^{-x}*ln e*-1\)

dy/dx=\(\displaystyle \L\ (x^{3})(e^{-x}*ln e*-1)+(3x^{2})(e^{-x})\)

dy/dx=\(\displaystyle \L\ (x^{3})(-e^{-x})+(3x^{2})(e^{-x})\)

dy/dx=\(\displaystyle \L\ x^{2}e^{-x}(-x+3)\)

another question... the derivitive of ln e is \(\displaystyle \L\frac{1}{x}\) right?

 

[galactus]

That's correct. Good

BTW, the derivative of ln(e) is 0. Since ln(e)=1, it's a constant and the derivative is 0. If you meant the derivative of ln(x), yes, it's 1/x.

 

[wind]

That's correct. Good

BTW, the derivative of ln(e) is 0. Since ln(e)=1, it's a constant and the derivative is 0. If you meant the derivative of ln(x), yes, it's 1/x.

Thanks galactus