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Hello!

Here are some differentiation rules to follow when trying to find the derivative of any function:

  • [tex]\frac{d}{dx} (x^{p}) = Px^{P-1}[/tex]
  • [tex]\frac{d}{dx} (a^{x}) = a^{x}ln a[/tex]
  • [tex]\frac{d}{dx} (e^{x}) = e^{x}[/tex]
  • [tex]\frac{d}{dx} (constant) = 0[/tex]
  • [tex]\frac{d}{dx}(x) = 1[/tex]
  • [tex]\frac{d}{dx} = (c * f(x)) = c * f'(x)[/tex]
  • [tex]\frac{d}{dx} (f(x) + g(x)) = f'(x) + g'(x)[/tex]
  • [tex]\frac{d}{dx} (f(x) - g(x)) = f'(x) - g'(x)[/tex]
  • [tex]\frac{d}{dx} (f(x) * g(x))= f(x) * g'(x) + g(x) * f'(x)[/tex]
  • [tex]\frac{d}{dx} (\frac{f(x)}{g(x)}) = \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^{2}}[/tex]

Now, let's find the derivative of these functions.

a). f(x) = 3 → f'(x) = 0

If you are differentiating a number, then the derivative is zero.

b). f(x) = a^x → f'(x) = a^x㏑(a)

c). f(x) = e^x → f'(x) = e^x

Looking at this, your first instinct would be to include ln(e). But, the natural log of e is ALWAYS equal to one. So, you would multiply e^x by 1, which is equal to e^x.

d). 2x^5 + 8x^3 - 3x → f'(x) = 10x^4 + 24x^2 - 3

With this, you multiply the power by the coefficient, and the power is reduced by 1. Also, the derivative of a variable is also equal to one.

2x^5: 5 x 2 = 10, 5 - 1 = 4 | 8x^3: 3 x 8 = 24, 3 - 1 = 2

Final answers:

  1. f'(x) = 0
  2. f'(x) = a^x㏑(a)
  3. f'(x) = e^x
  4. f'(x) = 10x^4 + 24x^2 - 3
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