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The differential of a polynomial term like `Ax^n` is `nAx^(n-1)` ` `
If `f(x)=x^4-2x^3+x^2`` `
Completing the basic math we get `f'(x)=4x^3-6x^2+2x`
Looking at the graphs of the original function (black) and it's derivative (blue), we see:
- For X values less than 0, the slope of the original function (black) is negative (as x values increase towards 0 from the left, the function decreases in value), therefore it's derivative (blue) is negative.
- From x=0 to x=0.5, the slope of the original function becomes positive (the value of the function rises as x increases), therefore the derivative (blue) is positive for this duration.
- From x=0.5 to x=1, the slope of the original function (black) is again negative as shown by the derivative (blue) again being below the x axis.
- For x values above 1, the original function again has a positive slope, resulting in a positive derivative.
Given f(x) = x^4 - 2x^3 +x^2
By differentiating on both sides we get
df(x)/dx = d/dx (x^4) - d/dx (2x^3) +d/dx (x^2)
f'(x) = 4 x^3 - 6 x^2 + 2 x
Positive sided graph is f(x) and both sided graph is f'(x)
we know at any point on f(x),f'(x) determines the slope of f(x) at that point.
f(x) = x^4 - 2x^3 +x^2
f'(x) = 4x^3 - 6x^2 + 2x
Note: The derivative at any point is the slope of the tangent to the curve at that point.
Both the graphs are plotted on the graphs,
As we see the graphs the f'(x) graph is positive when f(x) has a positive sloping tangent, when f(x) is increasing.
f '(x) is negative when f(x) has a negative sloping tangent, when f(x) is decreasing.
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