Derivative of f(x) = ln{(x^(4)) - 4/x}

[ladyazpy]

find the derivative of the function!

1. f(x) = ln{(x^(4)) - 4/x}

Qoutient Rule

={(x) (4x^(3)) - x^(4)-4} / {x} {1/x^(4)-4}

=4x^(4)+4x^(4) /x^(2) {x/x^(4)-4}

ans.

=8x^(4)/x^(2) times x/x^(4)-4

 

[arthur ohlsten]

y=ln[x^4-4/x] rewrite
y=ln[ x^5-4]/x]

let z=[ x^5-4]/x

y=ln z
dy/dx = 1/z dz/dx

1/z= x/[x^5-4]

dz/dx = x[5x^4]-[x^5-4] all over x^2
dz/dx= 5x^5-x^5+4 all over x^2
dz/dx= 4[x^5+1]/x^2

dy/dx = {x/[x^5-4]}{4[x^5+1]/x^2
dy/dx = 4[x^5+1] / {x[x^5-4]} answer

 

[ladyazpy]

im wondering wher x^5 came from

 

[daon]

im wondering wher x^5 came from

Your original post had \(\displaystyle \ln\( x^4 - \frac{4}{x} \)\). All arthur did was combine: \(\displaystyle \ln\( \frac{x^5 - 4}{x} \)\).

...Although I get the feeling you originally meant \(\displaystyle \ln\(\frac{x^4-4}{x}\)\).

This problem would have been much easier splitting apart the log with the following rule: \(\displaystyle \log\(\frac{a}{b}\) = \log$a$-\log$b$\).

Let $\displaystyle g(x)=x^4-4$ and $\displaystyle h(x) = x$.
Then $\displaystyle g'(x) = 4x^3$ and $\displaystyle h'(x) = 1$.

You have \(\displaystyle f(x) = \ln\(\frac{g(x)}{h(x)}\)\).

Using the rule, \(\displaystyle f(x) = \ln\(g(x)\) - \ln$h(x)$\).

Now, use the rule for differentiating the log: $\displaystyle f'(x) = \frac{g'(x)}{g(x)} - \frac{h'(x)}{h(x)}$

Now just substitute: $\displaystyle f'(x) = \frac{4x^3}{x^4-4} - \frac{1}{x}$.

At this point you're done but you could combine the fractions together if you'd like.

 

[soroban]

Re: Derivative

Hello, ladyazpy!

Assuming that the function is: $\displaystyle \:f(x)\;=\;\ln\left(\frac{x^4\,-\,4} {x}\right)$

. . we have: $\displaystyle \:f(x) \;=\;\ln(x^4\,-\,4)\,-\,\ln(x)$


Then: $\displaystyle \:f'(x)\;=\;\frac{4x^3}{x^4\,-\,4}\,-\,\frac{1}{x}$

. . which simplifies to: $\displaystyle \:f(x) \;=\;\frac{3x^4\,+\,4}{x(x^4\,-\,4)}$