Math Problem Statement

Calculate the derivative of the following function.

yequals=tangent left parenthesis e Superscript x Baseline right parenthesistanex

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Part 1

StartFraction dy Over dx EndFractiondydxequals=enter your response here

Solution

To calculate the derivative of the given function:

y=tan(ex)y = \tan(e^x)

We will use the chain rule, which states that:

ddx[f(g(x))]=f(g(x))g(x)\frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x)

Here, f(x)=tan(x)f(x) = \tan(x) and g(x)=exg(x) = e^x.

Step-by-step process:

  1. Differentiate the outer function (tangent): The derivative of tan(u)\tan(u) is sec2(u)\sec^2(u). So, applying this to tan(ex)\tan(e^x):

    ddx(tan(ex))=sec2(ex)\frac{d}{dx} \left( \tan(e^x) \right) = \sec^2(e^x)

  2. Differentiate the inner function (exponential): The derivative of exe^x is simply exe^x.

  3. Apply the chain rule: Now, multiply the derivative of the outer function by the derivative of the inner function:

    dydx=sec2(ex)ex\frac{dy}{dx} = \sec^2(e^x) \cdot e^x

Thus, the derivative of y=tan(ex)y = \tan(e^x) is:

dydx=sec2(ex)ex\frac{dy}{dx} = \sec^2(e^x) \cdot e^x

Would you like further details or have any questions about this?


Here are 5 related questions for further practice:

  1. What is the derivative of y=tan(2x)y = \tan(2x)?
  2. How do you differentiate y=tan1(ex)y = \tan^{-1}(e^x)?
  3. What is the second derivative of y=tan(ex)y = \tan(e^x)?
  4. How does the chain rule apply to differentiating trigonometric functions?
  5. What is the derivative of y=sec(ex)y = \sec(e^x)?

Tip: Always remember the chain rule when you have a function composed of multiple functions!

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Math Problem Analysis

Mathematical Concepts

Calculus
Differentiation
Chain Rule
Trigonometric Functions

Formulas

Derivative of tan(x): sec^2(x)
Chain Rule: d/dx [f(g(x))] = f'(g(x)) * g'(x)

Theorems

Chain Rule
Derivative of Exponential Functions

Suitable Grade Level

Grades 11-12 / College Level