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What is the derivative of y=(x^2√(1+x))/((1+x^2)^3/2))

User Mfollett
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2 Answers

2 votes

Answer:

Explanation:


f(x)={x^2√(1+x)\over{(1+x^2)^{3\over{2}}}}\\\\f^\prime (x)=\frac{{{(x^2√(1+x))^\prime (1+x^2)^{(3)/(2)}+ (x^2√(1+x))}((1+x^2)^{(3)/(2)}})^\prime}{((1+x^2)^(3)/(2))^2}}}


(x^2 √(1+x))^\prime=2x√(1+x)+ (x^2)/(2√(1+x))


((1+x^2)^{(3)/(2)})^\prime=(3)/(2)(1+x^2)^{(1)/(2)} (2x)=3x(1+x^2)^{(1)/(2)}


f^\prime(x)=\frac{\{(1+x^2)^{(3)/(2)}\}(2x√(1+x)+(x^2)/(2√(1+x)))+(x^2 √(1+x))(3x√(1+x^2))}{(1+x^2)^(3)}}


√(1+x^2)=√((1+x)(1-x))


f^\prime (x)=x√(1+x)~\frac{(1+x^2)^{(3)/(2)}(2+(x^2)/(2(1+x)))+3x^2√(1-x)}{(1+x^2)^3}

User Shravan Kumar
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5 votes

Explanation:

Maple calculator used.........................

What is the derivative of y=(x^2√(1+x))/((1+x^2)^3/2))-example-1
User Intrepion
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6.3k points