Math Problem Statement

Find the equation of the tangent line to the curve y = 5ln(x^5 * e^(x^4)) at the point (1,5).

Solution

The question in the image asks to find the equation of the tangent line to the curve:

y=5ln(x5ex4)y = 5 \ln \left( x^5 e^{x^4} \right)

at the point (1,5)(1, 5), and the answer should be in the form y=mx+by = mx + b.

Steps to Solve:

  1. Simplify the function: y=5ln(x5ex4)y = 5 \ln \left( x^5 e^{x^4} \right) Using properties of logarithms: y=5[ln(x5)+ln(ex4)]y = 5 \left[ \ln \left( x^5 \right) + \ln \left( e^{x^4} \right) \right] y=5(5ln(x)+x4)y = 5 \left( 5 \ln(x) + x^4 \right) y=25ln(x)+5x4y = 25 \ln(x) + 5x^4

  2. Differentiate to find the slope: To find the slope mm, we need the derivative of yy with respect to xx: dydx=ddx(25ln(x)+5x4)\frac{dy}{dx} = \frac{d}{dx} \left( 25 \ln(x) + 5x^4 \right) Derivative of 25ln(x)25 \ln(x) is 25x\frac{25}{x}, and derivative of 5x45x^4 is 20x320x^3: dydx=25x+20x3\frac{dy}{dx} = \frac{25}{x} + 20x^3

  3. Evaluate the derivative at x=1x = 1: dydxx=1=251+20(1)3=25+20=45\frac{dy}{dx} \bigg|_{x=1} = \frac{25}{1} + 20(1)^3 = 25 + 20 = 45 So, the slope m=45m = 45.

  4. Use the point-slope form: The equation of the tangent line can be written as: yy1=m(xx1)y - y_1 = m(x - x_1) where m=45m = 45, (x1,y1)=(1,5)(x_1, y_1) = (1, 5). Substituting: y5=45(x1)y - 5 = 45(x - 1) y5=45x45y - 5 = 45x - 45 y=45x40y = 45x - 40

Final Answer:

The equation of the tangent line is: y=45x40y = 45x - 40

Would you like further details or have any other questions?


Here are some related questions you might find useful:

  1. How do logarithmic differentiation rules apply to other complex functions?
  2. How can the product and chain rules be combined for more complex derivatives?
  3. What is the geometric interpretation of a tangent line to a curve?
  4. How does the derivative at a point determine the slope of the tangent line?
  5. What are other ways to find tangent lines without differentiating directly?

Tip: Always simplify the logarithmic and exponential expressions before differentiating, as it can make the process much easier and less error-prone.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Tangent Line
Logarithmic Functions

Formulas

Derivative of ln(u): d/dx[ln(u)] = u'/u
Product rule: d/dx[uv] = u'v + uv'
Chain rule: d/dx[f(g(x))] = f'(g(x)) * g'(x)

Theorems

The equation of a tangent line at a point (x_1, y_1) is y - y_1 = m(x - x_1)
The slope of the tangent line is the derivative of the function at that point

Suitable Grade Level

Grades 11-12