Derive the function x^6+(sin(x))/(8x^2+1) with respect to x

\frac{d}{dx}\left(x^6+\frac{\sin\left(x\right)}{8x^2+1}\right)

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Answer

$\frac{\left(1+8x^2\right)\cos\left(x\right)-16x\sin\left(x\right)}{\left(1+8x^2\right)^2}+6x^{5}$

Step by step solution

Problem

$\frac{d}{dx}\left(x^6+\frac{\sin\left(x\right)}{8x^2+1}\right)$
1

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

$\frac{d}{dx}\left(\frac{\sin\left(x\right)}{1+8x^2}\right)+\frac{d}{dx}\left(x^6\right)$
2

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

$\frac{d}{dx}\left(\frac{\sin\left(x\right)}{1+8x^2}\right)+6x^{5}$
3

Applying the quotient rule which states that if $f(x)$ and $g(x)$ are functions and $h(x)$ is the function defined by ${\displaystyle h(x) = \frac{f(x)}{g(x)}}$, where ${g(x) \neq 0}$, then ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

$\frac{\left(1+8x^2\right)\frac{d}{dx}\left(\sin\left(x\right)\right)-\frac{d}{dx}\left(1+8x^2\right)\sin\left(x\right)}{\left(1+8x^2\right)^2}+6x^{5}$
4

The derivative of the sine of a function is equal to the cosine of that function times the derivative of that function, in other words, if ${f(x) = \sin(x)}$, then ${f'(x) = \cos(x)\cdot D_x(x)}$

$\frac{\left(1+8x^2\right)\cos\left(x\right)-\frac{d}{dx}\left(1+8x^2\right)\sin\left(x\right)}{\left(1+8x^2\right)^2}+6x^{5}$
5

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

$\frac{\left(1+8x^2\right)\cos\left(x\right)-\left(\frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(8x^2\right)\right)\sin\left(x\right)}{\left(1+8x^2\right)^2}+6x^{5}$
6

The derivative of the constant function is equal to zero

$\frac{\left(1+8x^2\right)\cos\left(x\right)-\left(0+\frac{d}{dx}\left(8x^2\right)\right)\sin\left(x\right)}{\left(1+8x^2\right)^2}+6x^{5}$
7

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

$\frac{\left(1+8x^2\right)\cos\left(x\right)-\left(0+8\frac{d}{dx}\left(x^2\right)\right)\sin\left(x\right)}{\left(1+8x^2\right)^2}+6x^{5}$
8

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

$\frac{\left(1+8x^2\right)\cos\left(x\right)-\left(0+8\cdot 2x\right)\sin\left(x\right)}{\left(1+8x^2\right)^2}+6x^{5}$
9

Multiply $2$ times $8$

$\frac{\left(1+8x^2\right)\cos\left(x\right)-\left(0+16x\right)\sin\left(x\right)}{\left(1+8x^2\right)^2}+6x^{5}$
10

$x+0=x$, where $x$ is any expression

$\frac{\left(1+8x^2\right)\cos\left(x\right)-1\cdot 16x\sin\left(x\right)}{\left(1+8x^2\right)^2}+6x^{5}$
11

Multiply $16$ times $-1$

$\frac{\left(1+8x^2\right)\cos\left(x\right)-16x\sin\left(x\right)}{\left(1+8x^2\right)^2}+6x^{5}$

Answer

$\frac{\left(1+8x^2\right)\cos\left(x\right)-16x\sin\left(x\right)}{\left(1+8x^2\right)^2}+6x^{5}$

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Problem Analysis

Main topic:

Differential calculus

Time to solve it:

0.26 seconds

Views:

117