Question

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

Answer 1

Answer:

Step-by-step explanation:

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

$(x^2 \sqrt{1+x})^\prime=2x\sqrt{1+x}+ \frac{x^2}{2\sqrt{1+x}}$

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

$f^\prime(x)=\frac{\{(1+x^2)^{\frac{3}{2}}\}(2x\sqrt{1+x}+\frac{x^2}{2\sqrt{1+x}})+(x^2 \sqrt{1+x})(3x\sqrt{1+x^2})}{(1+x^2)^{3}}}$

$\sqrt{1+x^2}=\sqrt{(1+x)(1-x)}$

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

Answer 2

Step-by-step explanation:

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