Example Question - tangent line
Here are examples of questions we've helped users solve.
Determining the Slope of a Tangent to a Curve Involving Radicals
<p>Given \( f(x) = x^{\frac{7}{5}} + \frac{4}{\sqrt[5]{x^3}} \), to find the slope of the tangent line at any point \( x \), we need to find the derivative of \( f(x) \) with respect to \( x \).</p> <p>First, we rewrite the function in a more derivative-friendly form:</p> <p>\( f(x) = x^{\frac{7}{5}} + 4x^{-\frac{3}{5}} \)</p> <p>Now we differentiate \( f(x) \) with respect to \( x \) using the power rule \( \frac{d}{dx} x^n = nx^{n-1} \):</p> <p>\( f'(x) = \frac{7}{5}x^{\frac{7}{5}-1} - \frac{12}{5}x^{-\frac{3}{5}-1} \)</p> <p>Simplifying, we get:</p> <p>\( f'(x) = \frac{7}{5}x^{\frac{2}{5}} - \frac{12}{5}x^{-\frac{8}{5}} \)</p> <p>So, the slope of the tangent line to the curve at any point \( x \) is given by \( f'(x) \).</p>
Finding Tangent Line to Composition of Functions
Given \( g(x) = \frac{16}{x} \) and \( f(x) = -4 \sqrt{x} \), find the tangent line to \( g(f(x)) \) at \( x=4 \). Step 1: Compute \( f(4) \). \( f(4) = -4 \sqrt{4} = -4 \cdot 2 = -8 \) Step 2: Compute \( g(f(4)) \). \( g(-8) = \frac{16}{-8} = -2 \) Step 3: Compute \( f'(x) \). \( f'(x) = -4 \cdot \frac{1}{2}x^{-\frac{1}{2}} = -2x^{-\frac{1}{2}} \) Step 4: Compute \( f'(4) \). \( f'(4) = -2 \cdot 4^{-\frac{1}{2}} = -2 \cdot \frac{1}{2} = -1 \) Step 5: Compute \( g'(x) \). \( g'(x) = -16x^{-2} \) Step 6: Compute \( g'(-8) \). \( g'(-8) = -16(-8)^{-2} = -16 \cdot \frac{1}{64} = -\frac{1}{4} \) Step 7: Apply chain rule to find derivative of \( g(f(x)) \) at \( x=4 \). \( (g(f(x)))' = g'(f(x)) \cdot f'(x) \) \( (g(f(4)))' = g'(-8) \cdot f'(4) = -\frac{1}{4} \cdot (-1) = \frac{1}{4} \) Step 8: Write the equation of the tangent line. \( y = mx + b \) \( m = \frac{1}{4} \) \( y = \frac{1}{4}x + b \) Step 9: Use point \( (f(4), g(f(4))) = (4, -2) \) to solve for \( b \). \( -2 = \frac{1}{4} \cdot 4 + b \) \( -2 = 1 + b \) \( b = -3 \) Step 10: Final equation of the tangent line. \( y = \frac{1}{4}x - 3 \)
Angle Calculation Using Tangents and Congruent Segments
The image depicts a geometric diagram involving a circle with a tangent line. The information provided in the text accompanying the image is as follows: - \( YW \) is a tangent to the circle at point \( X \). - \( UV \) is congruent to \( VX \) (which means \( UV \) equals \( VX \)). - Angle \( VXW \) measures 50 degrees. To solve for the angle \( UXY \), we need to use the properties of a tangent to a circle and the fact that UV equals VX. When a line is tangent to a circle, it is perpendicular to the radius at the point of tangency. Therefore, angle \( UXY \), being the angle between the tangent line \( YW \) and the radius \( UX \), is a right angle or 90 degrees. Therefore, the value of angle \( UXY \) is 90 degrees.
Geometric Problem: Tangent Line and Circle
The image contains a geometric problem involving a circle with a tangent line. The problem statement reads: "In the diagram, the center of the circle is \( O \) and \( OT = 23 \). Calculate \( DT \)." Let's analyze the diagram. There's a triangle \( OAT \) with a right angle at \( A \), because the radius of a circle is perpendicular to the tangent at the point of tangency. Given that \( OT \) (the radius) is \( 23 \) units long, we are dealing with the Pythagorean theorem to find \( DT \). Since \( OA = OT \) (both are radii of the same circle), \( OA \) is also \( 23 \) units long. Now we have: \[ OA^2 + AD^2 = OT^2 \] To find \( DT \), we need to realize that \( DT = AD \). Therefore, we actually need to find \( AD \). Given that \( OA = OT = 23 \) units, by substituting the values in the Pythagorean theorem, we get: \[ 23^2 + AD^2 = 23^2 \] \[ 529 + AD^2 = 529 \] \[ AD^2 = 529 - 529 \] \[ AD^2 = 0 \] \[ AD = 0 \] Thus, \( DT = AD = 0 \). This indicates that point \( D \) coincides with point \( A \), and the length of segment \( DT \) is \( 0 \) units.
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