Exercise$\frac{d}{dx}\left(\left(x^2+3\right)^{5x-1}\right)$
Step-by-step Solution
1
To derive the function $\left(x^2+3\right)^{\left(5x-1\right)}$, use the method of logarithmic differentiation. First, assign the function to $y$, then take the natural logarithm of both sides of the equation
$y=\left(x^2+3\right)^{\left(5x-1\right)}$
2
Apply natural logarithm to both sides of the equality
$\ln\left(y\right)=\ln\left(\left(x^2+3\right)^{\left(5x-1\right)}\right)$
3
Using the power rule of logarithms: $\log_a(x^n)=n\cdot\log_a(x)$
$\ln\left(y\right)=\left(5x-1\right)\ln\left(x^2+3\right)$
4
Derive both sides of the equality with respect to $x$
$\frac{d}{dx}\left(\ln\left(y\right)\right)=\frac{d}{dx}\left(\left(5x-1\right)\ln\left(x^2+3\right)\right)$
5
Apply the product rule for differentiation: $(f\cdot g)'=f'\cdot g+f\cdot g'$, where $f=5x-1$ and $g=\ln\left(x^2+3\right)$
$\frac{d}{dx}\left(\ln\left(y\right)\right)=\frac{d}{dx}\left(5x-1\right)\ln\left(x^2+3\right)+\left(5x-1\right)\frac{d}{dx}\left(\ln\left(x^2+3\right)\right)$
Intermediate steps
6
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 $\displaystyle f'(x)=\frac{a'}{a}$
$\frac{1}{y}\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(5x-1\right)\ln\left(x^2+3\right)+\left(5x-1\right)\frac{1}{x^2+3}\frac{d}{dx}\left(x^2+3\right)$
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7
The derivative of the linear function is equal to $1$
$\frac{y^{\prime}}{y}=\frac{d}{dx}\left(5x-1\right)\ln\left(x^2+3\right)+\left(5x-1\right)\frac{1}{x^2+3}\frac{d}{dx}\left(x^2+3\right)$
Intermediate steps
8
The derivative of a sum of two or more functions is the sum of the derivatives of each function
$\frac{y^{\prime}}{y}=\frac{d}{dx}\left(5x\right)\ln\left(x^2+3\right)+\left(5x-1\right)\frac{1}{x^2+3}\frac{d}{dx}\left(x^2+3\right)$
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Intermediate steps
9
The derivative of a sum of two or more functions is the sum of the derivatives of each function
$\frac{y^{\prime}}{y}=\frac{d}{dx}\left(5x\right)\ln\left(x^2+3\right)+\left(5x-1\right)\frac{1}{x^2+3}\frac{d}{dx}\left(x^2\right)$
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Intermediate steps
10
The derivative of the linear function times a constant, is equal to the constant
$\frac{y^{\prime}}{y}=5\frac{d}{dx}\left(x\right)\ln\left(x^2+3\right)+\left(5x-1\right)\frac{1}{x^2+3}\frac{d}{dx}\left(x^2\right)$
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11
The derivative of the linear function is equal to $1$
$\frac{y^{\prime}}{y}=5\ln\left(x^2+3\right)+\left(5x-1\right)\frac{1}{x^2+3}\frac{d}{dx}\left(x^2\right)$
Intermediate steps
12
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$\frac{y^{\prime}}{y}=5\ln\left(x^2+3\right)+2\left(5x-1\right)\frac{1}{x^2+3}x$
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Intermediate steps
13
Multiply the fraction by the term
$\frac{y^{\prime}}{y}=5\ln\left(x^2+3\right)+\frac{2\left(5x-1\right)x}{x^2+3}$
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14
Multiply both sides of the equation by $y$
$y^{\prime}=\left(5\ln\left(x^2+3\right)+\frac{2\left(5x-1\right)x}{x^2+3}\right)y$
15
Substitute $y$ for the original function: $\left(x^2+3\right)^{\left(5x-1\right)}$
$y^{\prime}=\left(5\ln\left(x^2+3\right)+\frac{2\left(5x-1\right)x}{x^2+3}\right)\left(x^2+3\right)^{\left(5x-1\right)}$
16
The derivative of the function results in
$\left(5\ln\left(x^2+3\right)+\frac{2\left(5x-1\right)x}{x^2+3}\right)\left(x^2+3\right)^{\left(5x-1\right)}$
Intermediate steps
17
Simplify the derivative
$\left(5x^2\ln\left(x^2+3\right)+15\ln\left(x^2+3\right)+10x^2-2x\right)\left(x^2+3\right)^{\left(5x-2\right)}$
Explain this step further
Final answer to the exercise $\left(5x^2\ln\left(x^2+3\right)+15\ln\left(x^2+3\right)+10x^2-2x\right)\left(x^2+3\right)^{\left(5x-2\right)}$