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Considering the function `f(x) = (x - 1)/(x + 1),` you need to use quotient rule to evaluate derivative of the function, such that:
`f'(x) = ((x - 1)'(x + 1) - (x - 1)(x + 1)')/((x + 1)^2)`
`f'(x) = (x + 1 - x + 1)/((x + 1)^2)`
Reducing duplicate terms yields:
`f'(x) = 2/((x + 1)^2)`
Considering a function `f(x) = sqrt(x^2)` , you need to use chain rule to evaluate `f'(x)` , such that:
`f'(x) = (sqrt x^2)'*(x^2)'`
`f'(x) = 1/(2sqrt x^2)*(2x)`
Reducing duplicate factors yields:
`f'(x) = x/(sqrt x^2) => f'(x) = x/|x| => f'(x) = 1`
Hence, you need to use different rules of differentiation, considering the nature of the given function whose derivative you need to evaluate.
To determine the derivative of a function of a function, we'll have to apply the chain rule.
If u is a function of v and v is a function of x, then we can say that u is the function of the function v.
We'll write in this manner:
du/dx = (du/dv)*(dv/dx)
The first step is to find the function v (usually is the function inside the brackets or the argument of the trigonometric functions, or the argument of the logarithmic functions, or the expression under the square root, etc.)
Then, we'll re-write the function u in terms of v and we'll differentiate u with respect to v.
We'll re-write the results with respect to x.
u = (x^2 + 5x)^3
We'll have to differentiate u with respect to x:
du/dx = (d/dx)(x^2 + 5x)^3
We'll substitute the expression inside the brackets by v.
v = x^2 + 5x
u = v^3
To apply the chain rule, we'll have to differentiate u with respect to v:
du/dv = (v^3)'
du/dv = 3v^2
Now, we'll differentiate v with respect to x:
dv/dx = (x^2 + 5x)'
dv/dx = 2x + 5
du/dx = (du/dv)(dv/dx)
du/dx = 3[(x^2 + 5x)^2]*(2x + 5)
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