Math Problem Statement

Derivate 34x\exponentialE5x3^{4x}-\exponentialE^{5x}

Solution

To find the derivative of the function:

f(x)=34xe5xf(x) = 3^{4x} - e^{5x}

we'll differentiate each term separately.

Step 1: Differentiate 34x3^{4x}

To differentiate 34x3^{4x}, we use the general formula for the derivative of an exponential function with a base other than ee: ddx(au)=auln(a)u\frac{d}{dx} \left( a^{u} \right) = a^{u} \ln(a) \cdot u' where a=3a = 3 and u=4xu = 4x, so u=4u' = 4.

Thus, ddx(34x)=34xln(3)4=4ln(3)34x\frac{d}{dx} \left( 3^{4x} \right) = 3^{4x} \ln(3) \cdot 4 = 4 \ln(3) \cdot 3^{4x}

Step 2: Differentiate e5xe^{5x}

To differentiate e5xe^{5x}, we apply the chain rule: ddx(e5x)=e5x5=5e5x\frac{d}{dx} \left( e^{5x} \right) = e^{5x} \cdot 5 = 5 e^{5x}

Step 3: Combine the Results

Now, combining both derivatives, we have: f(x)=4ln(3)34x5e5xf'(x) = 4 \ln(3) \cdot 3^{4x} - 5 e^{5x}

Final Answer

f(x)=4ln(3)34x5e5xf'(x) = 4 \ln(3) \cdot 3^{4x} - 5 e^{5x}

Would you like further details on any step, or do you have any questions?

Related Questions

  1. How do we apply the chain rule in more complex functions?
  2. What is the derivative of axa^{x} where aea \neq e?
  3. How would the result change if the function were 34x+e5x3^{4x} + e^{5x} instead?
  4. How can we integrate 34x3^{4x} and e5xe^{5x}?
  5. What if we wanted to find the second derivative of this function?

Tip

For exponential functions with bases other than ee, remember to multiply by ln(base)\ln(\text{base}) after differentiating the exponent.

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Math Problem Analysis

Mathematical Concepts

Calculus
Differentiation
Exponential Functions

Formulas

Derivative of a^u: d/dx (a^u) = a^u * ln(a) * u'
Derivative of e^u: d/dx (e^u) = e^u * u'

Theorems

Chain Rule

Suitable Grade Level

Grades 11-12