Question

How do you find the derivative of $y=\sqrt{9-x}$?

Answer 1

Hint: To differentiate the function $y=\sqrt{9-x}$, convert the function $y=\sqrt{9-x}$ into$y={{\left( 9-x \right)}^{\dfrac{1}{2}}}$. Now to differentiate $y={{\left( 9-x \right)}^{\dfrac{1}{2}}}$we can use the normal formula for differentiation i.e. \[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\]. If the outer element is differentiated then the inner element has to be differentiated.

Complete step by step solution:
We have the given function $y=\sqrt{9-x}.......\left( i \right)$.
Convert the function$y=\sqrt{9-x}$ into $y={{\left( 9-x \right)}^{\dfrac{1}{2}}}$. The root symbol is \[\dfrac{1}{2}\]power of the element.
Differentiating equation (i) we get,
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( \sqrt{9-x} \right)\]
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}{{\left( 9-x \right)}^{\dfrac{1}{2}}}......\left( ii \right)\]
Differentiate using the formula \[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\] in the equation (ii).
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2}{{\left( 9-x \right)}^{\dfrac{1}{2}-1}}\cdot \dfrac{d}{dx}\left( 9-x \right)\]
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2}{{\left( 9-x \right)}^{-\dfrac{1}{2}}}\cdot \dfrac{d}{dx}\left( 9-x \right)\]
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2}{{\left( 9-x \right)}^{-\dfrac{1}{2}}}\cdot -1\]
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{-{{\left( 9-x \right)}^{-\dfrac{1}{2}}}}{2}......\left( iii \right)\]
Now in the equation (iii) convert \[-\dfrac{1}{2}\]power into the root over form.
\[\Rightarrow \dfrac{dy}{dx}=-\dfrac{1}{2\sqrt{9-x}}........\left( iv \right)\]
Thus, we have obtained the differential of the given function $y=\sqrt{9-x}$ as \[\dfrac{dy}{dx}=-\dfrac{1}{2\sqrt{9-x}}\].
Hence, the derivative of the function $y=\sqrt{9-x}$ is \[-\dfrac{1}{2\sqrt{9-x}}\].

Note:
While differentiating a function we should always keep in mind that we know the formula for differentiation i.e. \[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\]. There is also an alternate way to solve this derivative. If a function is \[y=\sqrt{x}\] then its derivative will be \[\dfrac{dy}{dx}=\dfrac{1}{2\sqrt{x}}\]. So this could be used for this function too but since instead of \[x\] there is \[9-x\]. So apply the same formula but further integrate \[9-x\].
So the alternate form will be given as:
$\begin{align}
 & \dfrac{dy}{dx}=\dfrac{d}{dx}\left( \sqrt{9-x} \right) \\
 & =\dfrac{1}{2\sqrt{9-x}}\cdot \dfrac{d}{dx}\left( 9-x \right) \\
 & =\dfrac{1}{2\sqrt{9-x}}\cdot -1 \\
 & =-\dfrac{1}{2\sqrt{9-x}}
\end{align}$
Hence we have got the result. This answer is exactly the same as the previous method but the technique is different. The previous technique requires the understanding of the concept and it does not require the memorization of the formula. In differentiation always try to solve using the concept and use the formula wherever necessary. This method is more efficient because it takes less time.