Question

If \(y=e^{4x}(92)^x\); then \(\frac{dy}{dx}\) is equal to
(\(\log_e x=\log x\))

• A. \(4x\,e^{4x}\cdot 92^{x-1}\)
• B. \(e^{4x}92^{x-1}[368+x]\)
• C. \(e^{4x}(92)^x[4+\log 92]\)
• D. \(e^{4x}92^x\log 92\)

Hint:

Use the product rule. \(\frac{d}{dx}92^x=92^x\log 92\).

Answer 1

The Correct Option is C

Step 1: Understand the function.
The function \(y=e^{4x}\cdot 92^{x}\) is a product of two exponential terms. The variable \(x\) sits in the exponent in both, so the ordinary power rule \(\frac{d}{dx}x^n=nx^{n-1}\) does not apply.

Step 2: Recall the formulas.
1. \(\frac{d}{dx}e^{kx}=k\,e^{kx}\).
2. \(\frac{d}{dx}a^{x}=a^{x}\log a\) for a constant \(a>0\).
3. Product rule: \((uv)'=u'v+uv'\).

Step 3: Differentiate the product.
Let \(u=e^{4x}\) and \(v=92^x\). Then \(u'=4e^{4x}\) and \(v'=92^x\log 92\).
\[ \frac{dy}{dx}=4e^{4x}\cdot 92^x+e^{4x}\cdot 92^x\log 92 \]

Step 4: Take out the common factor.
Both terms contain \(e^{4x}92^x\).
\[ \frac{dy}{dx}=e^{4x}(92)^x\,[\,4+\log 92\,] \]

Step 5: Check option 1.
Option 1 treats \(92^x\) like \(x^{92}\) and brings down a factor \(x\). That is the wrong rule, so it is wrong.

Step 6: Check option 2.
Option 2 also uses \(92^{x-1}\), which comes from the power rule again. It is wrong.

Step 7: Check option 4.
Option 4 differentiates only \(92^x\) and forgets the derivative of \(e^{4x}\). The term with 4 is missing, so it is wrong. Option 3 has both terms.

Final Answer:
The derivative is \(e^{4x}(92)^x[4+\log 92]\), which is option 3. \[ \boxed{e^{4x}(92)^x\,[4+\log 92]} \]