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y(x)=x^ four (8ln^2x-4lnx+ one)
y(x) equally x to the power of 4(8ln squared x minus 4lnx plus 1)
y(x) equally x to the power of four (8ln squared x minus 4lnx plus one)
y(x)=x4(8ln2x-4lnx+1)
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yx=x^48ln^2x-4lnx+1
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y(x)=x^4(8ln^2x+4lnx+1)
y(x)=x^4(8ln^2x-4lnx-1)

The derivative of the function/ y(x)=x^4(8ln^2x-4lnx+1)

Derivative of y(x)=x^4(8ln^2x-4lnx+1)

The solution

You have entered [src]

 4 /     2                  \
x *\8*log (x) - 4*log(x) + 1/

$$x^{4} \left(\left(8 \log{\left(x \right)}^{2} - 4 \log{\left(x \right)}\right) + 1\right)$$

x^4*(8*log(x)^2 - 4*log(x) + 1)

Detail solution

Apply the product rule:

$\frac{d}{d x} f{\left(x \right)} g{\left(x \right)} = f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}$

$f{\left(x \right)} = x^{4}$ ; to find $\frac{d}{d x} f{\left(x \right)}$ :
1. Apply the power rule: $x^{4}$ goes to $4 x^{3}$
$g{\left(x \right)} = \left(8 \log{\left(x \right)}^{2} - 4 \log{\left(x \right)}\right) + 1$ ; to find $\frac{d}{d x} g{\left(x \right)}$ :
1. Differentiate $\left(8 \log{\left(x \right)}^{2} - 4 \log{\left(x \right)}\right) + 1$ term by term:
  1. Differentiate $8 \log{\left(x \right)}^{2} - 4 \log{\left(x \right)}$ term by term:
    1. The derivative of a constant times a function is the constant times the derivative of the function.
      1. Let $u = \log{\left(x \right)}$ .
      2. Apply the power rule: $u^{2}$ goes to $2 u$
      3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \log{\left(x \right)}$ :
        
        The derivative of $\log{\left(x \right)}$ is $\frac{1}{x}$ .
        
        The result of the chain rule is:
        
        $\frac{2 \log{\left(x \right)}}{x}$
      So, the result is: $\frac{16 \log{\left(x \right)}}{x}$
    2. The derivative of a constant times a function is the constant times the derivative of the function.
      1. The derivative of $\log{\left(x \right)}$ is $\frac{1}{x}$ .
      So, the result is: $- \frac{4}{x}$
    The result is: $\frac{16 \log{\left(x \right)}}{x} - \frac{4}{x}$
  2. The derivative of the constant $1$ is zero.
  The result is: $\frac{16 \log{\left(x \right)}}{x} - \frac{4}{x}$
The result is: $x^{4} \left(\frac{16 \log{\left(x \right)}}{x} - \frac{4}{x}\right) + 4 x^{3} \left(\left(8 \log{\left(x \right)}^{2} - 4 \log{\left(x \right)}\right) + 1\right)$
Now simplify:

$32 x^{3} \log{\left(x \right)}^{2}$

The answer is:

$32 x^{3} \log{\left(x \right)}^{2}$

The graph

Plot the graph f(x)

Plot the graph f'(x)

The first derivative [src]

 4 /  4   16*log(x)\      3 /     2                  \
x *|- - + ---------| + 4*x *\8*log (x) - 4*log(x) + 1/
   \  x       x    /

$$x^{4} \left(\frac{16 \log{\left(x \right)}}{x} - \frac{4}{x}\right) + 4 x^{3} \left(\left(8 \log{\left(x \right)}^{2} - 4 \log{\left(x \right)}\right) + 1\right)$$

Simplify

The second derivative [src]

   2                                        
4*x *(28*log(x) + 12*(-1 + 2*log(x))*log(x))

$$4 x^{2} \left(12 \left(2 \log{\left(x \right)} - 1\right) \log{\left(x \right)} + 28 \log{\left(x \right)}\right)$$

Simplify

The third derivative [src]

8*x*(8 + 52*log(x) + 12*(-1 + 2*log(x))*log(x))

$$8 x \left(12 \left(2 \log{\left(x \right)} - 1\right) \log{\left(x \right)} + 52 \log{\left(x \right)} + 8\right)$$

Simplify

The graph

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Derivative of y(x)=x^4(8ln^2x-4lnx+1)

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The solution