Answer:
[tex]\frac{1}{10\sqrt{\frac{x}{5} -7} }.[/tex]
Step-by-step explanation:
First of all, refer to the chain rule of derivation.
Chain rule says the following: [tex]\frac{d}{dx}(\sqrt{x+a} ) =\frac{d}{du}(\sqrt{u} )*\frac{d}{dx}(x+a)[/tex], where [tex]u=x+a[/tex].
Let's find the derivative.
Step 1. Write the expression.
[tex]\frac{d}{dx} (\sqrt{\frac{x}{5} -7} )[/tex]
Step 2. Choose a value for u and rewrite the expression.
[tex]u=\frac{x}{5} -7\\ \\\frac{d}{du} (\sqrt{u} )*\frac{d}{dx} (\frac{x}{5} -7)[/tex]
Step 3. Find the derivatives of each element.
a. For the u term.
[tex]\frac{d}{du} (\sqrt{u} )=\\ \\\frac{d}{du} (u^{\frac{1}{2} })=\\ \\ \frac{1}{2} u^{\frac{1}{2} -1}= \\ \\ \frac{1}{2} u^{-\frac{1}{2}}= \\ \\\frac{1}{2\sqrt{u} }= \\\\\frac{1}{2\sqrt{\frac{x}{5} -7} }[/tex]
b. For the x term.
[tex]\frac{d}{dx} (\frac{x}{5}-7 )=\\ \\\frac{1}{5}-0=\\\\ \frac{1}{5}[/tex]
Step 4. Multiply the term to finish applying the chain rule.
[tex]\frac{1}{2\sqrt{\frac{x}{5} -7} }*\frac{1}{5}= \\ \\\frac{1}{10\sqrt{\frac{x}{5} -7} }[/tex]
Answer:
[tex]\dfrac{\text{d}y}{\text{d}x}= \dfrac{1}{10 \sqrt{\frac{x}{5}-7}}[/tex]
Step-by-step explanation:
Given equation:
[tex]y=\sqrt{\dfrac{x}{5}-7}[/tex]
To differentiate the given equation, use the chain rule.
[tex]\boxed{\begin{minipage}{5.4 cm}\underline{Chain Rule for Differentiation}\\\\If $y=f(u)$ and $u=g(x)$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=\dfrac{\text{d}y}{\text{d}u}\times \dfrac{\text{d}u}{\text{d}x}$\\\end{minipage}}[/tex]
[tex]\textsf{Let } \: y=\sqrt{u} \: \textsf{ where } \: u=\dfrac{x}{5}-7[/tex]
Differentiate the two parts separately:
[tex]\begin{aligned}y & = \sqrt{u}\\y&=u^{\frac{1}{2}}\\\implies \dfrac{\text{d}y}{\text{d}u} & =\dfrac{1}{2} \cdot u^{\frac{1}{2}-1}\\& =\dfrac{1}{2}u^{-\frac{1}{2}}\\& = \dfrac{1}{2u^{\frac{1}{2}}}\\& = \dfrac{1}{2\sqrt{u}}\end{aligned}[/tex]
[tex]\begin{aligned}u & = \dfrac{1}{5}x-7\\\implies \dfrac{\text{d}u}{\text{d}x} & = \dfrac{1}{5}\end{aligned}[/tex]
Put everything back into the chain rule formula:
[tex]\begin{aligned}\dfrac{\text{d}y}{\text{d}x} & =\dfrac{\text{d}y}{\text{d}u}\times \dfrac{\text{d}u}{\text{d}x}\\\\\implies \dfrac{\text{d}y}{\text{d}x} & = \dfrac{1}{2\sqrt{u}} \times \dfrac{1}{5}\\& = \dfrac{1}{10\sqrt{u}}\\& = \dfrac{1}{10 \sqrt{\frac{x}{5}-7}}\end{aligned}[/tex]
Differentiation Rules
[tex]\boxed{\begin{minipage}{4.4cm}\underline{Differentiating $x^n$}\\\\If $y=x^n$, then $\dfrac{\text{d}y}{\text{d}x}=nx^{n-1}$\\\end{minipage}}[/tex]
[tex]\boxed{\begin{minipage}{4.4 cm}\underline{Differentiating $ax$}\\\\If $y=ax$, then $\dfrac{\text{d}y}{\text{d}x}=a$\\\end{minipage}}[/tex]
[tex]\boxed{\begin{minipage}{4.4 cm}\underline{Differentiating a constant}\\\\If $y=a$, then $\dfrac{\text{d}y}{\text{d}x}=0$\\\end{minipage}}[/tex]
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