Find the derivative of xe^(3x-5)+ln(x-4)

\frac{d}{dx}\left(x e^{\left(3x-5\right)}+\ln\left(x-4\right)\right)

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Answer

$\frac{1}{x-4}+3xe^{\left(3x-5\right)}+e^{\left(3x-5\right)}$

Step by step solution

Problem

$\frac{d}{dx}\left(x e^{\left(3x-5\right)}+\ln\left(x-4\right)\right)$
1

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

$\frac{d}{dx}\left(\ln\left(x-4\right)\right)+\frac{d}{dx}\left(xe^{\left(3x-5\right)}\right)$
2

Apply the product rule for differentiation: $(f\cdot g)'=f'\cdot g+f\cdot g'$, where $f=x$ and $g=e^{\left(3x-5\right)}$

$\frac{d}{dx}\left(\ln\left(x-4\right)\right)+x\frac{d}{dx}\left(e^{\left(3x-5\right)}\right)+e^{\left(3x-5\right)}\cdot\frac{d}{dx}\left(x\right)$
3

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

$\frac{d}{dx}\left(\ln\left(x-4\right)\right)+x\frac{d}{dx}\left(e^{\left(3x-5\right)}\right)+1e^{\left(3x-5\right)}$
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 $\displaystyle f'(x)=\frac{a'}{a}$

$\frac{1}{x-4}\cdot\frac{d}{dx}\left(x-4\right)+x\frac{d}{dx}\left(e^{\left(3x-5\right)}\right)+1e^{\left(3x-5\right)}$
5

Applying the derivative of the exponential function

$\frac{1}{x-4}\cdot\frac{d}{dx}\left(x-4\right)+1x\frac{d}{dx}\left(3x-5\right)e^{\left(3x-5\right)}+1e^{\left(3x-5\right)}$
6

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

$\frac{1}{x-4}\left(\frac{d}{dx}\left(-4\right)+\frac{d}{dx}\left(x\right)\right)+1x\left(\frac{d}{dx}\left(-5\right)+\frac{d}{dx}\left(3x\right)\right)e^{\left(3x-5\right)}+1e^{\left(3x-5\right)}$
7

The derivative of the constant function is equal to zero

$\frac{1}{x-4}\left(0+\frac{d}{dx}\left(x\right)\right)+1x\left(0+\frac{d}{dx}\left(3x\right)\right)e^{\left(3x-5\right)}+1e^{\left(3x-5\right)}$
8

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

$\left(0+1\right)\left(\frac{1}{x-4}\right)+1x\left(0+\frac{d}{dx}\left(3x\right)\right)e^{\left(3x-5\right)}+1e^{\left(3x-5\right)}$
9

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

$\left(0+1\right)\left(\frac{1}{x-4}\right)+1x\left(0+3\frac{d}{dx}\left(x\right)\right)e^{\left(3x-5\right)}+1e^{\left(3x-5\right)}$
10

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

$\left(0+1\right)\left(\frac{1}{x-4}\right)+\left(0+1\cdot 3\right)\cdot 1xe^{\left(3x-5\right)}+1e^{\left(3x-5\right)}$
11

Add the values $1$ and $0$

$1\left(\frac{1}{x-4}\right)+\left(0+1\cdot 3\right)\cdot 1xe^{\left(3x-5\right)}+1e^{\left(3x-5\right)}$
12

Multiply $3$ times $1$

$1\left(\frac{1}{x-4}\right)+\left(0+3\right)\cdot 1xe^{\left(3x-5\right)}+1e^{\left(3x-5\right)}$
13

Add the values $3$ and $0$

$1\left(\frac{1}{x-4}\right)+3\cdot 1xe^{\left(3x-5\right)}+1e^{\left(3x-5\right)}$
14

Multiply $1$ times $3$

$1\left(\frac{1}{x-4}\right)+3xe^{\left(3x-5\right)}+1e^{\left(3x-5\right)}$
15

Any expression multiplied by $1$ is equal to itself

$\frac{1}{x-4}+3xe^{\left(3x-5\right)}+e^{\left(3x-5\right)}$

Answer

$\frac{1}{x-4}+3xe^{\left(3x-5\right)}+e^{\left(3x-5\right)}$

Problem Analysis

Main topic:

Differential calculus

Time to solve it:

0.23 seconds

Views:

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