d/dx (x⁴)(2x^-2 +3x^-3)
Answers
Step-by-step explanation:
How do I find "dy/dx" with the equation "y = 3x^4 + 2x^2 + 10"?
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Mage Chartreux
Answered 4 years ago
dy/dx = -12x^3 + 4x + 0
d/dx of y is dy/dx
d/dx of -3x^4 is, according to the power rule, -12x^3. The power rule dictates that one simply drops the exponent and makes it a coefficient of the variable and then subtracts one from the exponent: d/dx of x^n = nx^(n-1).
d/dx of 2x^2 is 4x (power rules again)
d/dx of 10 is 0; d/dx of a constant is 0.
In poorly formatted mathematical notation:
y = -3x^4 + 2x^2 + 10 — Original function
dy/dx = d/dx(-3x^4 + 2x^2 + 10) — Apply d/dx to both sides
dy/dx = d/dx(-3x^4) + d/dx(2x^2) + d/dx(10) — Apply d/dx to each term in the polynomial
dy/dx = -3 * d/dx(x^4) + 2 * d
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Profile photo for Guillermo Owen
Guillermo Owen
Answered 4 years ago
Let f(x) = 3x^4 + 2x^2 + 10
f(x+h) = 3(x^4 +4x^3 h + 6x^2 h^2 + 4xh^3 + h^4) +2(x^2 + 2xh + h^2) + 10
f(x+h) - f(x) = 12x^3 h + + 18x^2 h^2 + 12xh^3 + 3h^4) + 4xh +2h^2
(f(x+h)-f(x))/h = 12x^3 +18x^2 h + 12xh^2 +3h^3 +4x +2h
Take the limit of this last expression as h goes to 0. You get
f’(x) = 12x^3 + 4x
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