Question

f(x) = e^2x / x^5 + 1
f ' x = ?
This one seems reallly confusing

Answer 1

They can look quite confusing at first, but once you're comfortable with all the rules, they really aren't too bad. :)

In this equation, we can see that the second term on the right-hand side is just going to fall off (derivative of a constant being 0), so we can ignore it. The first term is a quotient of two functions, so we may use the quotient rule, or if we change the exponent of the denominator to a negative and bring it back up, it becomes easier for us to use the product rule.
\[e ^{2x}/x ^{5} = e^{2x}x^{-5}\]
The product rule applied to this will be...
\[(e^{2x} \times x^{-5})' = (e^{2x})'x^{-5} + e^{2x}(x^{-5})'\]

From the property of e^x and its derivatives being equal, we just use the chain rule. So, the derivative of e^(2x) is 2e^(2x). From there, it is mostly just simplification.

\[f'(x) = 2e^{2x}x^{-5} - 5e^{2x}x^{-6}\]

Of course, you may simplify further (return the exponents to positives), but this should be sufficient. :)

Answer 2

the answer should be in the form df/dx however that is

Answer 3

The f'(x) and df/dx mean the same thing. You can represent the derivative with either notation, so you just replace the "f'(x)" in the above equation with "df/dx".

Answer 4

how do you simplify that, the homework is being stuborn

Answer 5

Well, because you have negative exponents, you would place them back in the denominator with positive exponents (just a property of exponents).

\[df/dx = (2e^{2x}/x^{5}) - (5e^{2x}/x^{6})\]

(Sorry, these should be fractions, but I don't know how to add them with the equation thing here)

If you need to have one denominator, you just multiply the first term by (x/x)
\[df/dx = (2xe^{2x} - 5e^{2x})/x^{6}, = [(2x - 5)e^{2x}]/x^{6}\]

That should be completely simplified.

Answer 6

i hate my homework