Find the higher order derivative of x^33+ln(x+1)

\frac{d^2}{dx^2}\left(x^{33}+\ln\left(x+1\right)\right)

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Answer

$cosh\left(x\right)+1056x^{31}$

Step by step solution

Problem

$\frac{d^2}{dx^2}\left(x^{33}+\ln\left(x+1\right)\right)$
1

Rewriting the high order derivative

$\frac{d^{\left(2-1\right)}}{dx^{\left(2-1\right)}}\left(\frac{d}{dx}\left(\ln\left(1+x\right)+x^{33}\right)\right)$
2

The derivative of a sum of two functions is the sum of the derivatives of each function

$\frac{d^{\left(2-1\right)}}{dx^{\left(2-1\right)}}\left(\frac{d}{dx}\left(\ln\left(1+x\right)\right)+\frac{d}{dx}\left(x^{33}\right)\right)$
3

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$\frac{d^{\left(2-1\right)}}{dx^{\left(2-1\right)}}\left(\frac{d}{dx}\left(\ln\left(1+x\right)\right)+33x^{32}\right)$
4

The derivative of the natural logarithm of a function is equal to the derivative of the function divided by that function. If $f(x)=ln\:a$ (where $a$ is a function of $x$), then $\displaystyle f'(x)=\frac{a'}{a}$

$\frac{d^{\left(2-1\right)}}{dx^{\left(2-1\right)}}\left(\frac{1}{1+x}\cdot\frac{d}{dx}\left(1+x\right)+33x^{32}\right)$
5

The derivative of a sum of two functions is the sum of the derivatives of each function

$\frac{d^{\left(2-1\right)}}{dx^{\left(2-1\right)}}\left(\frac{1}{1+x}\left(\frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)\right)+33x^{32}\right)$
6

The derivative of the constant function is equal to zero

$\frac{d^{\left(2-1\right)}}{dx^{\left(2-1\right)}}\left(\frac{1}{1+x}\left(0+\frac{d}{dx}\left(x\right)\right)+33x^{32}\right)$
7

The derivative of the linear function is equal to $1$

$\frac{d^{\left(2-1\right)}}{dx^{\left(2-1\right)}}\left(\left(0+1\right)\left(\frac{1}{1+x}\right)+33x^{32}\right)$
8

Subtract the values $2$ and $-1$

$\frac{d^{1}}{dx^{1}}\left(\left(0+1\right)\left(\frac{1}{1+x}\right)+33x^{32}\right)$
9

Add the values $1$ and $0$

$\frac{d^{1}}{dx^{1}}\left(1\left(\frac{1}{1+x}\right)+33x^{32}\right)$
10

Any expression to the power of $1$ is equal to that same expression

$\frac{d}{dx}\left(1\left(\frac{1}{1+x}\right)+33x^{32}\right)$
11

Any expression multiplied by $1$ is equal to itself

$\frac{d}{dx}\left(\frac{1}{1+x}+33x^{32}\right)$
12

The derivative of a sum of two functions is the sum of the derivatives of each function

$\frac{d}{dx}\left(\frac{1}{1+x}\right)+\frac{d}{dx}\left(33x^{32}\right)$
13

The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function

$\frac{d}{dx}\left(\frac{1}{1+x}\right)+33\frac{d}{dx}\left(x^{32}\right)$
14

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$\frac{d}{dx}\left(\frac{1}{1+x}\right)+33\cdot 32x^{31}$
15

Taking the derivative of the hyperbolic sine

$\frac{d}{dx}\left(x\right)cosh\left(x\right)+33\cdot 32x^{31}$
16

The derivative of the linear function is equal to $1$

$1cosh\left(x\right)+33\cdot 32x^{31}$
17

Multiply $32$ times $33$

$1cosh\left(x\right)+1056x^{31}$
18

Any expression multiplied by $1$ is equal to itself

$cosh\left(x\right)+1056x^{31}$

Answer

$cosh\left(x\right)+1056x^{31}$

Problem Analysis

Main topic:

Differential calculus

Time to solve it:

0.23 seconds

Views:

92