Question

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

Answer 1

In this question we have the expression in the form of a composite function and since there is no direct formula to solve the question, we will use the chain rule which is denoted as $F'(x)=f'(g(x))g'(x)$, where the function $f(x)$ is the outer function and the function $g(x)$ is the inner function and the composite function is in the form of $f(g(x))$.

Complete step by step solution:
We have the given expression as $\sqrt{9-x}$.
We have to find the derivative of the given expression therefore; it can be written as:
$\Rightarrow \dfrac{d}{dx}\sqrt{9-x}$
Now since the term in the square root format, we can write it in the form of exponent as:
$\Rightarrow \dfrac{d}{dx}{{\left( 9-x \right)}^{\dfrac{1}{2}}}$
Now the expression in the form of a composite derivative therefore, we will use the chain rule which is: $F'(x)=f'(g(x))g'(x)$
In this case we have $g(x)=9-x$ .
Now we know that $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ therefore, on using the formula and differentiating, we get:
$\Rightarrow \dfrac{1}{2}{{\left( 9-x \right)}^{\dfrac{1}{2}-1}}\dfrac{d}{dx}\left( 9-x \right)$
On simplifying the term, we get:
$\Rightarrow \dfrac{1}{2}{{\left( 9-x \right)}^{-\dfrac{1}{2}}}\dfrac{d}{dx}\left( 9-x \right)$
Now since there is a negative exponent to the term, we can take the reciprocal and write it as:
$\Rightarrow \dfrac{1}{2}\times \dfrac{1}{{{\left( 9-x \right)}^{\dfrac{1}{2}}}}\dfrac{d}{dx}\left( 9-x \right)$
Now the term in the exponent can also be written in the form of square root as:
$\Rightarrow \dfrac{1}{2}\times \dfrac{1}{\sqrt{9-x}}\dfrac{d}{dx}\left( 9-x \right)$
On simplifying the terms, we get:
$\Rightarrow \dfrac{1}{2\sqrt{9-x}}\dfrac{d}{dx}\left( 9-x \right)$
Since the other two terms which are to be differentiated are in subtraction, we can split the derivative as:
$\Rightarrow \dfrac{1}{2\sqrt{9-x}}\left( \dfrac{d}{dx}9-\dfrac{d}{dx}x \right)$
Now we know that $\dfrac{d}{dx}k=0$ and $\dfrac{d}{dx}x=1$ therefore, on differentiating, we get:
$\Rightarrow \dfrac{1}{2\sqrt{9-x}}\left( 0-1 \right)$
On simplifying, we get:
$\Rightarrow \dfrac{-1}{2\sqrt{9-x}}$, which is the required solution.
Note: It is to be remembered that that $\dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2\sqrt{x}}$ should be remembered as a general result or a formula whenever its differentiation is to be found. It is to be remembered that chain rule is used only when the expression is in the form of a composite function, which means it is in the form of $f(g(x))$.