y = (-9e^7x)/(5x+3)
y' = -9(35x^2+21x+3)*e^7x / (5x+3)^20
Can someone please help me find the derivative of the following:
y = (-9e^7x) / (5x+3)
Thank you!
5 answers
Use the quotient rule:
d(u/v)=(vdu-udv)/v²
u=(-9e^(7x))
du/dx=-63e^7x
dv/dx=5
so
d(u/v)=((5x+3)(-63e^(7x))+9e^(7x)(5))/(5x+3)^2
=-9(35x+16)e^(7x)/(5x+3)²
d(u/v)=(vdu-udv)/v²
u=(-9e^(7x))
du/dx=-63e^7x
dv/dx=5
so
d(u/v)=((5x+3)(-63e^(7x))+9e^(7x)(5))/(5x+3)^2
=-9(35x+16)e^(7x)/(5x+3)²
typo
There is typo
I do not know where I got power 20
-9( 35x+ 16 ) * e^7x /(5x+3)^2
There is typo
I do not know where I got power 20
-9( 35x+ 16 ) * e^7x /(5x+3)^2
Thank you guys!
You're welcome!
1 answer