Math Problem Statement

Find the equation of the tangent line to the graph of the function f(x) = sin(9√x) at the point (π^2, 0).

Solution

We are asked to find the equation of the tangent line to the graph of the function:

f(x)=sin(9x)f(x) = \sin\left(9 \sqrt{x}\right)

at the point (π2,0)\left(\pi^2, 0\right).

Step 1: Find the derivative of f(x)f(x)

To get the equation of the tangent line, we need to compute the derivative f(x)f'(x), which gives us the slope of the tangent line at a given point.

Given f(x)=sin(9x)f(x) = \sin(9 \sqrt{x}), we will apply the chain rule for differentiation:

  • The outer function is sin(u)\sin(u), where u=9xu = 9 \sqrt{x}.
  • The derivative of sin(u)\sin(u) is cos(u)u\cos(u) \cdot u'.

Now, we need to find the derivative of u=9x=9x1/2u = 9 \sqrt{x} = 9 x^{1/2}. Using the power rule, the derivative is:

u=912x1/2=92xu' = 9 \cdot \frac{1}{2} x^{-1/2} = \frac{9}{2 \sqrt{x}}

Now, applying the chain rule:

f(x)=cos(9x)92xf'(x) = \cos(9 \sqrt{x}) \cdot \frac{9}{2 \sqrt{x}}

Step 2: Compute the slope at x=π2x = \pi^2

To find the slope of the tangent line at x=π2x = \pi^2, we substitute x=π2x = \pi^2 into the derivative f(x)f'(x):

f(π2)=cos(9π2)92π2f'(\pi^2) = \cos(9 \sqrt{\pi^2}) \cdot \frac{9}{2 \sqrt{\pi^2}}

Since π2=π\sqrt{\pi^2} = \pi, we simplify the expression:

f(π2)=cos(9π)92πf'(\pi^2) = \cos(9 \pi) \cdot \frac{9}{2 \pi}

We know that cos(9π)=1\cos(9 \pi) = -1 because cos(θ)\cos(\theta) has a period of 2π2\pi and 9π9\pi corresponds to an odd multiple of π\pi. Therefore:

f(π2)=192π=92πf'(\pi^2) = -1 \cdot \frac{9}{2 \pi} = -\frac{9}{2 \pi}

So, the slope of the tangent line at x=π2x = \pi^2 is 92π-\frac{9}{2 \pi}.

Step 3: Equation of the tangent line

The general form of the equation of a line is:

yy1=m(xx1)y - y_1 = m (x - x_1)

where mm is the slope, and (x1,y1)(x_1, y_1) is the point of tangency. In our case, (x1,y1)=(π2,0)(x_1, y_1) = (\pi^2, 0), and the slope m=92πm = -\frac{9}{2 \pi}. Substituting these values:

y0=92π(xπ2)y - 0 = -\frac{9}{2 \pi} (x - \pi^2)

Simplifying:

y=92π(xπ2)y = -\frac{9}{2 \pi} (x - \pi^2)

Thus, the equation of the tangent line is:

y=92π(xπ2)y = -\frac{9}{2 \pi} (x - \pi^2)


Would you like more details on any part of the solution?

Here are some related questions:

  1. How is the chain rule used in differentiation?
  2. What is the geometric significance of a tangent line?
  3. How do you find the cosine of an angle like 9π9\pi?
  4. Why is the derivative important for finding the slope of a tangent line?
  5. How does the power rule apply to square root functions?

Tip: When differentiating composite functions, always identify the inner and outer functions clearly to apply the chain rule properly.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Calculus
Differentiation
Tangent Line
Chain Rule

Formulas

f'(x) = cos(9√x) * (9 / (2√x))
Equation of tangent line: y - y1 = m(x - x1)

Theorems

Chain Rule in Differentiation

Suitable Grade Level

Grades 11-12