find slope of tangent line to a curve of 3/sqrt(x)
To find the slope of the tangent line to the curve of \(f(x) = \frac{3}{\sqrt{x}}\), we need to find the derivative of the function \(f(x)\).
Using the power rule for differentiation, we have:
\[
\begin{aligned}
f'(x) &= \frac{d}{dx} \left(\frac{3}{\sqrt{x}}\right) \\
&= 3 \cdot \frac{d}{dx} \left(x^{-1/2}\right) \\
&= 3 \cdot \left(-\frac{1}{2}\right) \cdot x^{-1/2 - 1} \\
&= -\frac{3}{2}x^{-3/2}.
\end{aligned}
\]
Therefore, the slope of the tangent line to the curve \(f(x) = \frac{3}{\sqrt{x}}\) is \(-\frac{3}{2}x^{-3/2}\).
To find the slope of the tangent line to the curve of 3/sqrt(x), we need to first find the derivative of the curve.
Step 1: Recall the power rule for differentiation. For any function y = ax^n, where a and n are constants, the derivative is given by dy/dx = anx^(n-1).
In this case, the function is y = 3/sqrt(x). We can rewrite it as y = 3x^(-1/2), where a = 3 and n = -1/2.
Step 2: Apply the power rule to find the derivative of y with respect to x. Taking the derivative of y = 3x^(-1/2), we have:
dy/dx = (-1/2) * 3 * x^(-1/2 - 1)
= (-1/2) * 3 * x^(-3/2)
= (-3/2) * x^(-3/2)
So, the derivative is dy/dx = (-3/2) * x^(-3/2).
Step 3: Simplify the derivative if possible. We can rewrite x^(-3/2) as 1/sqrt(x^3) by using the property x^(-n) = 1/x^n.
Therefore, the derivative becomes dy/dx = (-3/2) * 1/sqrt(x^3).
Step 4: To find the slope of the tangent line at a specific point, substitute the x-coordinate of the point into dy/dx. The result will be the slope of the tangent line at that point.
Let's say we want to find the slope of the tangent line at x = a, then the slope of the tangent line is given by substituting x = a into dy/dx:
slope = (-3/2) * 1/sqrt(a^3).
So, the slope of the tangent line to the curve of 3/sqrt(x) at x = a is (-3/2) * 1/sqrt(a^3).
To find the slope of the tangent line to the curve of the function \(f(x) = \frac{3}{\sqrt{x}}\), we need to differentiate the function using the rules of differentiation.
Let's proceed with finding the derivative of the function.
Step 1: Rewrite the function using exponent notation:
\(f(x) = 3x^{-1/2}\)
Step 2: Apply the power rule of differentiation:
The derivative of \(x^n\) is \(nx^{n-1}\).
Applying this rule to our function, we have:
\(f'(x) = (-1/2) \cdot 3x^{-1/2-1} = - \frac{3}{2} x^{-3/2}\)
Step 3: Simplify the expression:
\(f'(x) = - \frac{3}{2 \sqrt{x^3}}\)
The derivative \(f'(x) = - \frac{3}{2 \sqrt{x^3}}\) gives us the slope of the tangent line at any point on the curve.
use the formula of f(a+h)-f(a)/h
Certainly! Using the formula for the slope of a secant line, we can approximate the slope of the tangent line to the curve of the function \(f(x) = \frac{3}{\sqrt{x}}\) using the formula \(\frac{f(a+h) - f(a)}{h}\), where \(a\) is a specific value and \(h\) is a small change in \(x\).
Let's proceed with finding this slope at a specific point. Let's choose the value \(a\) to be \(a = 4\) for example.
Step 1: Determine \(f(a)\) and \(f(a+h)\).
\(f(a) = f(4) = \frac{3}{\sqrt{4}} = \frac{3}{2}\)
\(f(a+h) = f(4+h) = \frac{3}{\sqrt{4+h}}\)
Step 2: Substitute the values into the formula to calculate the slope of the secant line:
Slope of the secant line = \(\frac{f(a+h) - f(a)}{h} = \frac{\frac{3}{\sqrt{4+h}} - \frac{3}{2}}{h}\)
Step 3: Simplify the expression:
\(\frac{\frac{3}{\sqrt{4+h}} - \frac{3}{2}}{h} = \frac{3}{2} \cdot \frac{2 - \sqrt{4+h}}{\sqrt{4+h} \cdot h}\)
Therefore, the slope of the secant line to the curve \(f(x) = \frac{3}{\sqrt{x}}\) at the point \(a = 4\) is \(\frac{3}{2} \cdot \frac{2 - \sqrt{4+h}}{\sqrt{4+h} \cdot h}\), where \(h\) represents a small change in \(x\). This value approximates the slope of the tangent line to the curve at the point \(x = 4\).