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artyomya1
@artyomya1
August 2021
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посчитайте производную то-есть y'
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Indentuum
Правило дифференцирования сложной функции:
(f(g(x)))' = f'(g(x))*g'(x)
1) y = e^(6x^4)
y' = e^(6x^4) * 24x^3
2) y = root(5,2^(6x-5))
y' = 1/5root(5, (2^(6x-5))^4)) * 2^(6x-5) * ln2 * 6
3) y = e^x * arccosx
y' = e^x*arccosx - e^x/sqrt(1-x^2)
4) y = 5(ln(3x^2 + 1))^7
y' = 35(ln(3x^2+1))^6 * 1/(3x^2 +1) * 6x
5) y = 3log(2, (4-5x)^4)
y' = 3/((4-5x)^4 * ln2) * 4(4-5x)^3 * 5
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Answers & Comments
(f(g(x)))' = f'(g(x))*g'(x)
1) y = e^(6x^4)
y' = e^(6x^4) * 24x^3
2) y = root(5,2^(6x-5))
y' = 1/5root(5, (2^(6x-5))^4)) * 2^(6x-5) * ln2 * 6
3) y = e^x * arccosx
y' = e^x*arccosx - e^x/sqrt(1-x^2)
4) y = 5(ln(3x^2 + 1))^7
y' = 35(ln(3x^2+1))^6 * 1/(3x^2 +1) * 6x
5) y = 3log(2, (4-5x)^4)
y' = 3/((4-5x)^4 * ln2) * 4(4-5x)^3 * 5