Question

find the differential dy. y = √(15 + x²) dy = (b) evaluate dy for the given values of x and dx. x = 1, dx = - 0.2 dy =

Explanation:

Step1: Use chain - rule for differentiation

If $y = \sqrt{u}=u^{\frac{1}{2}}$ and $u = 15 + x^{2}$, then $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$ and $\frac{du}{dx}=2x$. By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$.

Step2: Calculate $\frac{dy}{dx}$

Substitute $u = 15 + x^{2}$ into $\frac{dy}{du}$ and multiply by $\frac{du}{dx}$:
$\frac{dy}{dx}=\frac{1}{2}(15 + x^{2})^{-\frac{1}{2}}\cdot2x=\frac{x}{\sqrt{15 + x^{2}}}$. Then $dy=\frac{x}{\sqrt{15 + x^{2}}}dx$.

Step3: Evaluate $dy$ for given $x$ and $dx$

Substitute $x = 1$ and $dx=-0.2$ into $dy=\frac{x}{\sqrt{15 + x^{2}}}dx$.
$dy=\frac{1}{\sqrt{15 + 1}}\times(-0.2)=\frac{1}{\sqrt{16}}\times(-0.2)=\frac{1}{4}\times(-0.2)=-0.05$.

Answer:

(a) $dy=\frac{x}{\sqrt{15 + x^{2}}}dx$
(b) $dy=-0.05$