Differentiate f(x) = 5𝑥³+7𝑥 2𝑥²−4𝑥+5 using quotient rule.
f′(x)=(5x3+7x2x2−4x+5)′==(5x3+7x)′(2x2−4x+5)−(5x3+7x)(2x2−4x+5)′(2x2−4x+5)2==(15x2+7)(2x2−4x+5)−(5x3+7x)(4x−4)(2x2−4x+5)2==10x4−40x3+61x2+35(2x2−4x+5)2f'(x)=\left(\frac{5x^3+7x}{2x^2-4x+5}\right)'=\\ {}=\frac{(5x^3+7x)'(2x^2-4x+5)-(5x^3+7x)(2x^2-4x+5)'}{(2x^2-4x+5)^2}=\\ {}=\frac{(15x^2+7)(2x^2-4x+5)-(5x^3+7x)(4x-4)}{(2x^2-4x+5)^2}=\\ {}=\frac{10x^4-40x^3+61x^2+35}{(2x^2-4x+5)^2}f′(x)=(2x2−4x+55x3+7x)′==(2x2−4x+5)2(5x3+7x)′(2x2−4x+5)−(5x3+7x)(2x2−4x+5)′==(2x2−4x+5)2(15x2+7)(2x2−4x+5)−(5x3+7x)(4x−4)==(2x2−4x+5)210x4−40x3+61x2+35
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