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y=x^2-28x+96lnx-5

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Identical expressions
y=x^ two -28x+96lnx- five
y equally x squared minus 28x plus 96lnx minus 5
y equally x to the power of two minus 28x plus 96lnx minus five
y=x2-28x+96lnx-5
y=x²-28x+96lnx-5
y=x to the power of 2-28x+96lnx-5
Similar expressions
y=x^2+28x+96lnx-5
y=x^2-28x+96lnx+5
y=x^2-28x-96lnx-5

The derivative of the function/ y=x^2-28x+96lnx-5

Derivative of y=x^2-28x+96lnx-5

The solution

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 2                       
x  - 28*x + 96*log(x) - 5

$$x^{2} - 28 x + 96 \log{\left(x \right)} - 5$$

d / 2                       \
--\x  - 28*x + 96*log(x) - 5/
dx

$$\frac{d}{d x} \left(x^{2} - 28 x + 96 \log{\left(x \right)} - 5\right)$$

Transform

Detail solution

Differentiate $x^{2} - 28 x + 96 \log{\left(x \right)} - 5$ term by term:
1. Apply the power rule: $x^{2}$ goes to $2 x$
2. The derivative of a constant times a function is the constant times the derivative of the function.
  1. The derivative of a constant times a function is the constant times the derivative of the function.
    1. Apply the power rule: $x$ goes to $1$
    So, the result is: $28$
  So, the result is: $-28$
3. 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: $\frac{96}{x}$
4. The derivative of the constant $\left(-1\right) 5$ is zero.
The result is: $2 x - 28 + \frac{96}{x}$

The answer is:

$2 x - 28 + \frac{96}{x}$

The graph

Plot the graph f(x)

Plot the graph f'(x)

The first derivative [src]

            96
-28 + 2*x + --
            x

$$2 x - 28 + \frac{96}{x}$$

Simplify

The second derivative [src]

  /    48\
2*|1 - --|
  |     2|
  \    x /

$$2 \cdot \left(1 - \frac{48}{x^{2}}\right)$$

Simplify

The third derivative [src]

192
---
  3
 x

$$\frac{192}{x^{3}}$$

Simplify

The graph

Derivative of y=x^2-28x+96lnx-5