Derive the function (3(1-1sin(x)))/(2cos(x)) with respect to x

\frac{d}{dx}\left(\frac{3\left(1-\sin\left(x\right)\right)}{2\cos\left(x\right)}\right)

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Answer

$\frac{6\left(1-\sin\left(x\right)\right)\sin\left(x\right)-6\cos\left(x\right)^2}{4\cos\left(x\right)^2}$

Step by step solution

Problem

$\frac{d}{dx}\left(\frac{3\left(1-\sin\left(x\right)\right)}{2\cos\left(x\right)}\right)$
1

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 ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

$\frac{2\frac{d}{dx}\left(3\left(1-\sin\left(x\right)\right)\right)\cos\left(x\right)-3\frac{d}{dx}\left(2\cos\left(x\right)\right)\left(1-\sin\left(x\right)\right)}{\left(2\cos\left(x\right)\right)^2}$
2

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

$\frac{2\cdot 3\cos\left(x\right)\frac{d}{dx}\left(1-\sin\left(x\right)\right)-3\cdot 2\left(1-\sin\left(x\right)\right)\frac{d}{dx}\left(\cos\left(x\right)\right)}{\left(2\cos\left(x\right)\right)^2}$
3

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

$\frac{2\cdot 3\cos\left(x\right)\frac{d}{dx}\left(1-\sin\left(x\right)\right)-3\cdot 2\left(-1\right)\left(1-\sin\left(x\right)\right)\sin\left(x\right)}{\left(2\cos\left(x\right)\right)^2}$
4

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

$\frac{2\cdot 3\cos\left(x\right)\left(\frac{d}{dx}\left(-\sin\left(x\right)\right)+\frac{d}{dx}\left(1\right)\right)-3\cdot 2\left(-1\right)\left(1-\sin\left(x\right)\right)\sin\left(x\right)}{\left(2\cos\left(x\right)\right)^2}$
5

The derivative of the constant function is equal to zero

$\frac{2\cdot 3\cos\left(x\right)\left(\frac{d}{dx}\left(-\sin\left(x\right)\right)+0\right)-3\cdot 2\left(-1\right)\left(1-\sin\left(x\right)\right)\sin\left(x\right)}{\left(2\cos\left(x\right)\right)^2}$
6

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

$\frac{2\cdot 3\cos\left(x\right)\left(0-\frac{d}{dx}\left(\sin\left(x\right)\right)\right)-3\cdot 2\left(-1\right)\left(1-\sin\left(x\right)\right)\sin\left(x\right)}{\left(2\cos\left(x\right)\right)^2}$
7

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 ${f'(x) = \cos(x)\cdot D_x(x)}$

$\frac{2\cdot 3\cos\left(x\right)\left(0-\cos\left(x\right)\right)-3\cdot 2\left(-1\right)\left(1-\sin\left(x\right)\right)\sin\left(x\right)}{\left(2\cos\left(x\right)\right)^2}$
8

Multiply $3$ times $2$

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

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

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

Multiply $-1$ times $6$

$\frac{6\left(1-\sin\left(x\right)\right)\sin\left(x\right)-6\cos\left(x\right)\cos\left(x\right)}{\left(2\cos\left(x\right)\right)^2}$
11

When multiplying exponents with same base you can add the exponents

$\frac{6\left(1-\sin\left(x\right)\right)\sin\left(x\right)-6\cos\left(x\right)^2}{\left(2\cos\left(x\right)\right)^2}$
12

The power of a product is equal to the product of it's factors raised to the same power

$\frac{6\left(1-\sin\left(x\right)\right)\sin\left(x\right)-6\cos\left(x\right)^2}{4\cos\left(x\right)^2}$

Answer

$\frac{6\left(1-\sin\left(x\right)\right)\sin\left(x\right)-6\cos\left(x\right)^2}{4\cos\left(x\right)^2}$

Problem Analysis

Main topic:

Differential calculus

Time to solve it:

0.34 seconds

Views:

134