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Y = 2x / ln2x
y=2x/ln2x
y'=(2ln2x-2x*(1/2x)*2)/(ln2x)²
=(2ln2x-2)/(ln2x)²
Y = e ^ (- 2x) * sin3x
y'=(e^(-2x))'*sin3x+e^(-2x)*(sin3x)'=-2e^(-2x)sin3x+e^(-2x)cos3x*3==-2e^(-2x)sin3x+3e^(-2x)cos3x=e^(-2x)*(3cos3x-2sin3x)
Derivation f (x) = (2x + 1) ^ 5
The answer is 10 * (2x + 1) ^ 4. Please set u = 2x + 1 before doing it
f'(x)=5(2x+1)^4*(2x+1)'
=102(2x+1)^4
F (- 2x) = e ^ x, f '(x) = derivative
f(-2x)=e^x
Then f (x) = e ^ (- X / 2)
So f '(x) = e ^ (- X / 2) * (- X / 2)'
=e^(-x/2)*(-1/2)
=e^(-x/2)/2
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