differentiate the function.\n$p(w)=\frac{2w^{2}-8w + 6}{sqrt{w}}$\n$p(w)=\n\nresources\nread it watch it…

Explanation:

Step1: Rewrite the function $P(w)$

Rewrite $\frac{2w^{2}-8w + 6}{\sqrt{w}}$ as $2w^{\frac{3}{2}}-8w^{\frac{1}{2}}+6w^{-\frac{1}{2}}$.

Step2: Apply the power - rule for differentiation

The power - rule states that if $y = x^{n}$, then $y^\prime=nx^{n - 1}$. For $y = 2w^{\frac{3}{2}}$, $y^\prime=2\times\frac{3}{2}w^{\frac{3}{2}-1}=3w^{\frac{1}{2}}$. For $y=-8w^{\frac{1}{2}}$, $y^\prime=-8\times\frac{1}{2}w^{\frac{1}{2}-1}=-4w^{-\frac{1}{2}}$. For $y = 6w^{-\frac{1}{2}}$, $y^\prime=6\times(-\frac{1}{2})w^{-\frac{1}{2}-1}=-3w^{-\frac{3}{2}}$. So $P^\prime(w)=3w^{\frac{1}{2}}-4w^{-\frac{1}{2}}-3w^{-\frac{3}{2}}$.

Step3: Rewrite the function $z$

Rewrite $z=\frac{A}{y^{11}}+Be^{y}$ as $z = Ay^{-11}+Be^{y}$.

Step4: Differentiate $z$ with respect to $y$

Using the power - rule for $Ay^{-11}$: If $y = Ay^{-11}$, then $y^\prime=A\times(-11)y^{-11 - 1}=-11Ay^{-12}$. Using the rule that the derivative of $e^{y}$ is $e^{y}$ for $Be^{y}$: If $y = Be^{y}$, then $y^\prime=Be^{y}$. So $z^\prime=-11Ay^{-12}+Be^{y}$.

Answer:

$P^\prime(w)=3w^{\frac{1}{2}}-4w^{-\frac{1}{2}}-3w^{-\frac{3}{2}}$ $z^\prime=-11Ay^{-12}+Be^{y}$