Can someone please help me find the derivative of the following:

y = (-9e^7x) / (5x+3)

Thank you!

y = (-9e^7x)/(5x+3)

y' = -9(35x^2+21x+3)*e^7x / (5x+3)^20

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

typo

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!

To find the derivative of the given function, we can use the quotient rule. The quotient rule states that if we have a function of the form f(x) = g(x) / h(x), the derivative of f(x) is given by:

f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / h(x)^2

Now let's apply the quotient rule to the given function, where g(x) = -9e^(7x) and h(x) = 5x + 3:

Step 1: Find g'(x)
g'(x) = d/dx(-9e^(7x))
= -9 * d/dx(e^(7x))
= -9 * (7e^(7x))
= -63e^(7x)

Step 2: Find h'(x)
h'(x) = d/dx(5x + 3)
= 5 * d/dx(x) + d/dx(3)
= 5 * 1 + 0
= 5

Step 3: Plug g'(x), h(x), and h'(x) into the quotient rule formula
f'(x) = (-63e^(7x) * (5x + 3) - (-9e^(7x)) * 5) / (5x + 3)^2

Simplifying further, we get:

f'(x) = (-63e^(7x) * (5x + 3) + 45e^(7x)) / (5x + 3)^2

So, the derivative of y = (-9e^7x) / (5x + 3) is f'(x) = (-63e^(7x) * (5x + 3) + 45e^(7x)) / (5x + 3)^2.