If $f(x) = \dfrac{x - 4}{2\sqrt{x}}$, then $f'(1)$ is (A) $\dfrac{5}{4}$ (B) $\dfrac{4}{5}$ (C) $1$ (D) $0$

> [!TLDR] > We find the derivative $f'(x)$ by first simplifying the function $f(x)$ into a sum of power terms, then applying the power rule. Finally, we substitute $x=1$ into $f'(x)$ to get the result $\boxed{\frac{5}{4}}$. The question asks for the value of $f'(1)$, which represents the derivative of the function $f(x)$ evaluated at $x=1$. Conceptually, $f'(1)$ gives the instantaneous rate of change of $f(x)$ at $x=1$, or geometrically, the slope of the tangent line to the curve $y=f(x)$ at the point where $x=1$. To find $f'(1)$, our strategy is to first determine the general derivative function $f'(x)$ and then substitute $x=1$ into that expression. 1. **Rewrite the function in a more suitable form for differentiation.** The given function is $f(x) = \dfrac{x - 4}{2\sqrt{x}}$. It is often easier to differentiate functions involving fractions if we can express them as a sum or difference of terms with $x$ raised to various powers. This avoids the need for the quotient rule, which can sometimes be more cumbersome. We can split the fraction: $f(x) = \dfrac{x}{2\sqrt{x}} - \dfrac{4}{2\sqrt{x}}$ Recall that $\sqrt{x} = x^{1/2}$. Substitute this into the expression: $f(x) = \dfrac{x^1}{2x^{1/2}} - \dfrac{4}{2x^{1/2}}$ Now, use the rules of exponents ($x^a/x^b = x^{a-b}$ and $1/x^a = x^{-a}$): For the first term: $\dfrac{x^1}{x^{1/2}} = x^{1 - 1/2} = x^{1/2}$. So, $\dfrac{1}{2}x^{1/2}$. For the second term: $\dfrac{4}{2x^{1/2}} = 2x^{-1/2}$. Thus, the simplified function is: $f(x) = \dfrac{1}{2}x^{1/2} - 2x^{-1/2}$ 2. **Differentiate $f(x)$ with respect to $x$.** We will use the power rule for differentiation, which states: > [!FORMULA] > If $g(x) = cx^n$, then $g'(x) = cnx^{n-1}$. Apply the power rule to each term in $f(x)$: * For the first term, $\dfrac{1}{2}x^{1/2}$: Here, $c = \dfrac{1}{2}$ and $n = \dfrac{1}{2}$. Its derivative is $\dfrac{1}{2} \cdot \dfrac{1}{2} x^{(1/2 - 1

If f(x) = dfracx - 42√(x), then f'(1) is (A) (5)/( – Class 11 Mathematics

NCERT Exemplar · Q69

f(x) = \dfrac{x - 4}{2\sqrt{x}}

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Concept understanding — Derivative at a Point

Derivative at a Point: From Intuition to Precision

Imagine you're driving a car. Your speedometer doesn't tell you your average speed over the whole trip — it tells you your speed right now, at this exact instant. That's the core idea of a derivative at a point: it measures how fast something is changing at a single moment.

The Intuition: Instantaneous Rate of Change

Let's start with something simpler. Suppose you drop a ball from a height. The distance it has fallen after $t$ seconds is given by $s(t) = 4.9t^2$ metres (ignoring air resistance).

If I ask you "how fast was the ball falling after exactly 2 seconds?", you can't just divide distance by time — that gives an average speed over an interval. You need the speed at $t=2$ , not between $t=1$ and $t=3$ .

Here's the trick: take a very small time interval around $t=2$ , say from $t=2$ to $t=2+h$ where $h$ is tiny. The average speed over that interval is:

\frac{s(2+h) - s(2)}{h}

If $h = 0.1$ , you get one number. If $h = 0.01$ , you get a slightly different number. As $h$ gets closer and closer to 0, these average speeds approach a single value — that's the instantaneous speed at $t=2$ .

Note

This "shrinking interval" idea is the heart of the derivative. We're not setting $h=0$ (that would give $\frac{0}{0}$ , which is meaningless). We're letting $h$ approach 0 and seeing what the ratio approaches.

The Precise Definition

For a function $f(x)$ , the derivative at a point $x = a$ is defined as:

f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}

provided this limit exists.

Let's break this down piece by piece:

$f(a+h) - f(a)$ is the change in the function's value when you move from $a$ to $a+h$ .
Dividing by $h$ gives the average rate of change over that interval.
Taking the limit as $h \to 0$ shrinks the interval to a single point, giving the instantaneous rate of change.

f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}

Geometric Interpretation

There's also a beautiful geometric meaning. The average rate of change $\frac{f(a+h) - f(a)}{h}$ is the slope of the secant line through the points $(a, f(a))$ and $(a+h, f(a+h))$ .

As $h \to 0$ , these two points get closer together, and the secant line approaches a line that just touches the curve at $x=a$ — the tangent line. So:

The derivative at a point equals the slope of the tangent line to the curve at that point.

A Concrete Example

Let's compute the derivative of $f(x) = x^2$ at $x = 3$ .

Using the definition:

f'(3) = \lim_{h \to 0} \frac{(3+h)^2 - 3^2}{h}

Expand $(3+h)^2 = 9 + 6h + h^2$ :

f'(3) = \lim_{h \to 0} \frac{9 + 6h + h^2 - 9}{h} = \lim_{h \to 0} \frac{6h + h^2}{h}

Factor $h$ :

f'(3) = \lim_{h \to 0} \frac{h(6 + h)}{h} = \lim_{h \to 0} (6 + h)

Since $h \to 0$ , this approaches $6$ .

Important

The derivative of $x^2$ at $x=3$ is 6. This means:

At $x=3$ , the function is increasing at a rate of 6 units per unit change in $x$ .
The tangent line to $y = x^2$ at $(3, 9)$ has slope 6.

Notation

You'll see several notations for "the derivative of $f$ at $x=a$ ":

$f'(a)$ — Lagrange notation (most common) …

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We find the derivative $f'(x)$ by first simplifying the function $f(x)$ into a sum of power terms, then applying the power rule. Finally, we substitute $x=1$ into $f'(x)$ to get the result $\boxed{\frac{5}{4}}$ .

The question asks for the value of $f'(1)$ , which represents the derivative of the function $f(x)$ evaluated at $x=1$ . Conceptually, $f'(1)$ gives the instantaneous rate of change of $f(x)$ at $x=1$ , or geometrically, the slope of the tangent line to the curve $y=f(x)$ at the point where $x=1$ .

To find $f'(1)$ , our strategy is to first determine the general derivative function $f'(x)$ and then substitute $x=1$ into that expression.

Rewrite the function in a more suitable form for differentiation.

The given function is $f(x) = \dfrac{x - 4}{2\sqrt{x}}$ .

It is often easier to differentiate functions involving fractions if we can express them as a sum or difference of terms with $x$ raised to various powers. This avoids the need for the quotient rule, which can sometimes be more cumbersome.

We can split the fraction:

$f(x) = \dfrac{x}{2\sqrt{x}} - \dfrac{4}{2\sqrt{x}}$

Recall that $\sqrt{x} = x^{1/2}$ . Substitute this into the expression:

$f(x) = \dfrac{x^1}{2x^{1/2}} - \dfrac{4}{2x^{1/2}}$

Now, use the rules of exponents ( $x^a/x^b = x^{a-b}$ and $1/x^a = x^{-a}$ ):

For the first term: $\dfrac{x^1}{x^{1/2}} = x^{1 - 1/2} = x^{1/2}$ . So, $\dfrac{1}{2}x^{1/2}$ .

For the second term: $\dfrac{4}{2x^{1/2}} = 2x^{-1/2}$ .

Thus, the simplified function is:

$f(x) = \dfrac{1}{2}x^{1/2} - 2x^{-1/2}$
Differentiate $f(x)$ with respect to $x$ .

We will use the power rule for differentiation, which states:

If $g(x) = cx^n$ , then $g'(x) = cnx^{n-1}$ .

Apply the power rule to each term in $f(x)$ :
- For the first term, $\dfrac{1}{2}x^{1/2}$ :
  
  Here, $c = \dfrac{1}{2}$ and $n = \dfrac{1}{2}$ .
  
  Its derivative is $\dfrac{1}{2} \cdot \dfrac{1}{2} x^{(1/2 - 1)} = \dfrac{1}{4} x^{-1/2}$ .
- For the second term, $-2x^{-1/2}$ :
  
  Here, $c = -2$ and $n = -\dfrac{1}{2}$ . …

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Q.If f(x)=x−42xf(x) = \dfrac{x - 4}{2\sqrt{x}}f(x)=2x​x−4​, then f′(1)f'(1)f′(1) is (A) 54\dfrac{5}{4}45​ (B) 45\dfrac{4}{5}54​ (C) 111 (D) 000

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Q.If $f(x) = \dfrac{x - 4}{2\sqrt{x}}$ , then $f'(1)$ is
(A) $\dfrac{5}{4}$
(B) $\dfrac{4}{5}$
(C) $1$
(D) $0$

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