Differentiation of exponents x

• cal.queen92
In summary, the process for finding the derivative of f(x) = 2sin(x) + (4x)^x involved using the product rule and the chain rule. The final answer is (2cos(3) + (12)^3)(12+(1/4)).
cal.queen92

Homework Statement

If f(x) = 2sin(x) + (4x)^x find ---> f ' (3)

Homework Equations

d/dx (e^u) = e^u du/dx

The Attempt at a Solution

Here is my attempt:

f(x) = 2sin(x) + e^(ln(4x)^x) = 2sin(x) + e^(xln4x) = d/dx (2sin(x)) + d/dx (e^(xln4x))

= 2cos(x) + e^(xln4x) d/dx (xln4x) = 2cos(x) + e^(ln(4x)^x)*(4x+(x/4x))

= 2cos(x) + ((4x)^x) * (4x+(1/4)) = (unsimplified) --> (2cos(3) + (12)^3)(12+(1/4)

But this answer is incorrect. What am i doing wrong?

cal.queen92 said:

Homework Statement

If f(x) = 2sin(x) + (4x)^x find ---> f ' (3)

Homework Equations

d/dx (e^u) = e^u du/dx

The Attempt at a Solution

Here is my attempt:

f(x) = 2sin(x) + e^(ln(4x)^x) = 2sin(x) + e^(xln4x) = d/dx (2sin(x)) + d/dx (e^(xln4x))

= 2cos(x) + e^(xln4x) d/dx (xln4x) = 2cos(x) + e^(ln(4x)^x)*(4x+(x/4x))

= 2cos(x) + ((4x)^x) * (4x+(1/4)) = (unsimplified) --> (2cos(3) + (12)^3)(12+(1/4)

But this answer is incorrect. What am i doing wrong?

I believe your error lies in the bold parts. Bold parts should be equal. But $\frac{d}{dx}(xln4x)\ne (4x+(x/4x)$. You should have a ln(4x) factor in there somewhere.

Last edited:
cal.queen92 said:

Homework Statement

If f(x) = 2sin(x) + (4x)^x find ---> f ' (3)

Homework Equations

d/dx (e^u) = e^u du/dx

The Attempt at a Solution

Here is my attempt:

f(x) = 2sin(x) + e^(ln(4x)^x) = 2sin(x) + e^(xln4x) = d/dx (2sin(x)) + d/dx (e^(xln4x))

= 2cos(x) + e^(xln4x) d/dx (xln4x) = 2cos(x) + e^(ln(4x)^x)*(4x+(x/4x))
Mistake at the end, above. When you differentiate x ln4x, you have to use the product rule and the chain rule.

d/dx(x ln(4x)) = ln(4x) + x * (1/(4x)) * 4 = ln(4x) + 1.
cal.queen92 said:
= 2cos(x) + ((4x)^x) * (4x+(1/4)) = (unsimplified) --> (2cos(3) + (12)^3)(12+(1/4)

But this answer is incorrect. What am i doing wrong?

That helps! Awsome! It worked out, thank you!

1. What is the product rule for differentiating exponents?

The product rule for differentiating exponents states that when differentiating a term with two or more exponents, the derivative is found by multiplying the original term by the sum of the exponents and then subtracting one from each exponent.

2. How do you differentiate a term with a negative exponent?

To differentiate a term with a negative exponent, rewrite the term as a fraction with a positive exponent in the denominator. Then, use the quotient rule to find the derivative.

3. Can you differentiate a term with a fractional exponent?

Yes, you can differentiate a term with a fractional exponent. To do so, use the power rule, which states that the derivative of x to the power of n is equal to n times x to the power of n-1.

4. Is there a specific rule for differentiating exponential functions?

Yes, there is a specific rule for differentiating exponential functions. The derivative of an exponential function with base a is equal to a times the function itself, ln(a). In other words, the derivative of a to the power of x is equal to a to the power of x times ln(a).

5. Can I use the chain rule to differentiate terms with exponents?

Yes, you can use the chain rule to differentiate terms with exponents. The chain rule allows you to differentiate functions within functions, such as a term with an exponent inside a trigonometric function. The rule states that the derivative of the outer function multiplied by the derivative of the inner function.

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