Find the derivative of (e^xcos(x))/(tan(x)+sin(x))e^xln(x)*-1

\frac{d}{dx}\left(\frac{e^x\cos\left(x\right)}{\tan\left(x\right)+\sin\left(x\right)}-e^x\cdot\ln\left(x\right)\right)

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Answer

$\frac{\left(\sin\left(x\right)+\tan\left(x\right)\right)\left(\cos\left(x\right)e^x-e^x\sin\left(x\right)\right)-\left(\cos\left(x\right)+\sec\left(x\right)^2\right)\cos\left(x\right)e^x}{\left(\sin\left(x\right)+\tan\left(x\right)\right)^2}-\left(\frac{1}{x}e^x+\ln\left(x\right)e^x\right)$

Step by step solution

Problem

$\frac{d}{dx}\left(\frac{e^x\cos\left(x\right)}{\tan\left(x\right)+\sin\left(x\right)}-e^x\cdot\ln\left(x\right)\right)$
1

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

$\frac{d}{dx}\left(-\ln\left(x\right)e^x\right)+\frac{d}{dx}\left(\frac{\cos\left(x\right)e^x}{\sin\left(x\right)+\tan\left(x\right)}\right)$
2

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

$\frac{d}{dx}\left(\frac{\cos\left(x\right)e^x}{\sin\left(x\right)+\tan\left(x\right)}\right)-\frac{d}{dx}\left(\ln\left(x\right)e^x\right)$
3

Apply the product rule for differentiation: $(f\cdot g)'=f'\cdot g+f\cdot g'$, where $f=e^x$ and $g=\ln\left(x\right)$

$\frac{d}{dx}\left(\frac{\cos\left(x\right)e^x}{\sin\left(x\right)+\tan\left(x\right)}\right)-\left(\frac{d}{dx}\left(\ln\left(x\right)\right)e^x+\ln\left(x\right)\frac{d}{dx}\left(e^x\right)\right)$
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 $\displaystyle f'(x)=\frac{a'}{a}$

$\frac{d}{dx}\left(\frac{\cos\left(x\right)e^x}{\sin\left(x\right)+\tan\left(x\right)}\right)-\left(\frac{1}{x}e^x\frac{d}{dx}\left(x\right)+\ln\left(x\right)\frac{d}{dx}\left(e^x\right)\right)$
5

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

$\frac{d}{dx}\left(\frac{\cos\left(x\right)e^x}{\sin\left(x\right)+\tan\left(x\right)}\right)-\left(1\left(\frac{1}{x}\right)e^x+\ln\left(x\right)\frac{d}{dx}\left(e^x\right)\right)$
6

Applying the derivative of the exponential function

$\frac{d}{dx}\left(\frac{\cos\left(x\right)e^x}{\sin\left(x\right)+\tan\left(x\right)}\right)-\left(1\left(\frac{1}{x}\right)e^x+\ln\left(x\right)e^x\right)$
7

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

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

Apply the product rule for differentiation: $(f\cdot g)'=f'\cdot g+f\cdot g'$, where $f=e^x$ and $g=\cos\left(x\right)$

$\frac{\left(\sin\left(x\right)+\tan\left(x\right)\right)\left(\frac{d}{dx}\left(\cos\left(x\right)\right)e^x+\cos\left(x\right)\frac{d}{dx}\left(e^x\right)\right)-\frac{d}{dx}\left(\sin\left(x\right)+\tan\left(x\right)\right)\cos\left(x\right)e^x}{\left(\sin\left(x\right)+\tan\left(x\right)\right)^2}-\left(1\left(\frac{1}{x}\right)e^x+\ln\left(x\right)e^x\right)$
9

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 $f'(x) = -\sin(x)\cdot D_x(x)$

$\frac{\left(\sin\left(x\right)+\tan\left(x\right)\right)\left(\cos\left(x\right)\frac{d}{dx}\left(e^x\right)-e^x\sin\left(x\right)\right)-\frac{d}{dx}\left(\sin\left(x\right)+\tan\left(x\right)\right)\cos\left(x\right)e^x}{\left(\sin\left(x\right)+\tan\left(x\right)\right)^2}-\left(1\left(\frac{1}{x}\right)e^x+\ln\left(x\right)e^x\right)$
10

Applying the derivative of the exponential function

$\frac{\left(\sin\left(x\right)+\tan\left(x\right)\right)\left(\cos\left(x\right)e^x-e^x\sin\left(x\right)\right)-\frac{d}{dx}\left(\sin\left(x\right)+\tan\left(x\right)\right)\cos\left(x\right)e^x}{\left(\sin\left(x\right)+\tan\left(x\right)\right)^2}-\left(1\left(\frac{1}{x}\right)e^x+\ln\left(x\right)e^x\right)$
11

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

$\frac{\left(\sin\left(x\right)+\tan\left(x\right)\right)\left(\cos\left(x\right)e^x-e^x\sin\left(x\right)\right)-\left(\frac{d}{dx}\left(\sin\left(x\right)\right)+\frac{d}{dx}\left(\tan\left(x\right)\right)\right)\cos\left(x\right)e^x}{\left(\sin\left(x\right)+\tan\left(x\right)\right)^2}-\left(1\left(\frac{1}{x}\right)e^x+\ln\left(x\right)e^x\right)$
12

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

$\frac{\left(\sin\left(x\right)+\tan\left(x\right)\right)\left(\cos\left(x\right)e^x-e^x\sin\left(x\right)\right)-\left(\cos\left(x\right)+\frac{d}{dx}\left(\tan\left(x\right)\right)\right)\cos\left(x\right)e^x}{\left(\sin\left(x\right)+\tan\left(x\right)\right)^2}-\left(1\left(\frac{1}{x}\right)e^x+\ln\left(x\right)e^x\right)$
13

The derivative of the tangent of a function is equal to secant squared of that function times the derivative of that function, in other words, if ${f(x) = tan(x)}$, then ${f'(x) = sec^2(x)\cdot D_x(x)}$

$\frac{\left(\sin\left(x\right)+\tan\left(x\right)\right)\left(\cos\left(x\right)e^x-e^x\sin\left(x\right)\right)-\left(\cos\left(x\right)+\frac{d}{dx}\left(x\right)\sec\left(x\right)^2\right)\cos\left(x\right)e^x}{\left(\sin\left(x\right)+\tan\left(x\right)\right)^2}-\left(1\left(\frac{1}{x}\right)e^x+\ln\left(x\right)e^x\right)$
14

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

$\frac{\left(\sin\left(x\right)+\tan\left(x\right)\right)\left(\cos\left(x\right)e^x-e^x\sin\left(x\right)\right)-\left(\cos\left(x\right)+1\sec\left(x\right)^2\right)\cos\left(x\right)e^x}{\left(\sin\left(x\right)+\tan\left(x\right)\right)^2}-\left(1\left(\frac{1}{x}\right)e^x+\ln\left(x\right)e^x\right)$
15

Any expression multiplied by $1$ is equal to itself

$\frac{\left(\sin\left(x\right)+\tan\left(x\right)\right)\left(\cos\left(x\right)e^x-e^x\sin\left(x\right)\right)-\left(\cos\left(x\right)+\sec\left(x\right)^2\right)\cos\left(x\right)e^x}{\left(\sin\left(x\right)+\tan\left(x\right)\right)^2}-\left(\frac{1}{x}e^x+\ln\left(x\right)e^x\right)$
16

Using the power rule of logarithms

$\frac{\left(\sin\left(x\right)+\tan\left(x\right)\right)\left(\cos\left(x\right)e^x-e^x\sin\left(x\right)\right)-\left(\cos\left(x\right)+\sec\left(x\right)^2\right)\cos\left(x\right)e^x}{\left(\sin\left(x\right)+\tan\left(x\right)\right)^2}-\left(\frac{1}{x}e^x+\ln\left(x^{\left(e^x\right)}\right)\right)$
17

Using the power rule of logarithms: $\log_a(x^n)=n\cdot\log_a(x)$

$\frac{\left(\sin\left(x\right)+\tan\left(x\right)\right)\left(\cos\left(x\right)e^x-e^x\sin\left(x\right)\right)-\left(\cos\left(x\right)+\sec\left(x\right)^2\right)\cos\left(x\right)e^x}{\left(\sin\left(x\right)+\tan\left(x\right)\right)^2}-\left(\frac{1}{x}e^x+\ln\left(x\right)e^x\right)$

Answer

$\frac{\left(\sin\left(x\right)+\tan\left(x\right)\right)\left(\cos\left(x\right)e^x-e^x\sin\left(x\right)\right)-\left(\cos\left(x\right)+\sec\left(x\right)^2\right)\cos\left(x\right)e^x}{\left(\sin\left(x\right)+\tan\left(x\right)\right)^2}-\left(\frac{1}{x}e^x+\ln\left(x\right)e^x\right)$

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Problem Analysis

Main topic:

Differential calculus

Time to solve it:

0.39 seconds

Views:

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