Question

question 2
if ( y(u) = int_{0}^{u^6} sqrt{11 + x^3} dx ) then
( y(u) = )
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question 3

Explanation:

Step1: Recall Leibniz Rule for derivatives

For $y(u)=\int_{a(u)}^{b(u)}f(x)dx$, $y'(u)=f(b(u))\cdot b'(u)-f(a(u))\cdot a'(u)$

Step2: Identify components of the integral

Here, $a(u)=0$, $b(u)=u^6$, $f(x)=\sqrt{11+x^3}$

Step3: Compute derivative of lower limit

$a'(u)=\frac{d}{du}(0)=0$

Step4: Compute derivative of upper limit

$b'(u)=\frac{d}{du}(u^6)=6u^5$

Step5: Substitute into Leibniz Rule

$y'(u)=f(u^6)\cdot 6u^5 - f(0)\cdot 0 = \sqrt{11+(u^6)^3}\cdot 6u^5$

Step6: Simplify the expression

$\sqrt{11+u^{18}}\cdot 6u^5 = 6u^5\sqrt{11+u^{18}}$

Answer:

$6u^5\sqrt{11+u^{18}}$