Question

Find the first derivative of the function " id="TexFormula1" title=" f(x) = \sqrt{4 - x} " alt=" f(x) = \sqrt{4 - x} " align="absmiddle" class="latex-formula">
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Answer 1

Answer:

$\bold{-\dfrac{1}{2\sqrt{4-x}}}$

Step-by-step explanation:

This can also be solved without using limits

Let  $y = \sqrt{4-x}$

The power rule of derivatives says that if we have a function

$y = a.x^n,$

then the first derivative given by $\frac{dy}{dx}$ or abbreviated as $y'$ is given by $n(ax^{n-1})$

We can use this in combination with the substitution rule of calculus

Let u = 4-x

Then we have $\frac{du}{dx} = \frac{d(4)}{dx} - \frac{d(x)}{dx} = 0 -1 = -1$

(First differential of a constant is 0 and first differential of x  is 1)

Substituting for u in the original expression $\sqrt{4-x}$ we get $\frac{dy}{du} = \frac{d}{du} (\sqrt{u}) = \frac{d}{du}(u^{1/2}) = \frac{1}{2} u^{\frac{1}{2} -1} = \frac{1}{2} u^{-1/2} =\frac{1}{2} \frac{1}{\sqrt{u} }$

The substitution rule states $\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$

So

$\frac{dy}{dx} = \frac{1}{2\sqrt{u} } . (-1)$

Substituting for u in terms of x we get
$\frac{dy}{dx} = \frac{1}{2\sqrt{4-x} } . (-1) = -\frac{1}{2\sqrt{4-x}}$

Answer 2

Step-by-step explanation:

Check the attachment for answer