+77 Let f(x) = 3x-2x2 . Use the definition of the derivative as the limit of the difference quocient to find f1(x).
Find the derivative of the following via implicit differentiation:
d/dx(f(x)) = d/dx(3 x-2 x^2)
The derivative of f(x) is f'(x):
f'(x) = d/dx(3 x-2 x^2)
Differentiate the sum term by term and factor out constants:
f'(x) = 3 d/dx(x)-2 d/dx(x^2)
The derivative of x is 1:
f'(x) = -2 (d/dx(x^2))+1 3
Use the power rule, d/dx(x^n) = n x^(n-1), where n = 2: d/dx(x^2) = 2 x:
f'(x) = 3-2 2 x
Simplify the expression:
f'(x) = 3-4 x
Expand the left hand side:
Answer: | f'(x) = 3 - 4x
Find the derivative of the following via implicit differentiation:
d/dx(f(x)) = d/dx(3 x-2 x^2)
The derivative of f(x) is f'(x):
f'(x) = d/dx(3 x-2 x^2)
Differentiate the sum term by term and factor out constants:
f'(x) = 3 d/dx(x)-2 d/dx(x^2)
The derivative of x is 1:
f'(x) = -2 (d/dx(x^2))+1 3
Use the power rule, d/dx(x^n) = n x^(n-1), where n = 2: d/dx(x^2) = 2 x:
f'(x) = 3-2 2 x
Simplify the expression:
f'(x) = 3-4 x
Expand the left hand side:
Answer: | f'(x) = 3 - 4x
+9673 \(f(x)= 3x-2x^2\)
\(f'(x)=\displaystyle\lim_{h\rightarrow 0}\dfrac{f(x+h)-f(x)}{h}\\ \;\;\;\;\;\;\;\;= \displaystyle\lim_{h\rightarrow 0}\dfrac{(3(x+h)-2(x+h)^2)-(3x-2x^2)}{h}\\ \;\;\;\;\;\;\;\;=\displaystyle\lim_{h\rightarrow 0} \dfrac{3x+3h-2x ^2-4xh-2h^2-3x+2x^2}{h}\\ \;\;\;\;\;\;\;\;=\displaystyle\lim_{h\rightarrow 0}\dfrac{3h-4xh-2h^2}h\\ \;\;\;\;\;\;\;\;=\displaystyle\lim_{h\rightarrow 0} (3-4x-2h)\\ \;\;\;\;\;\;\;\;= 3-4x\)
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