Derivative Problems with Exponential and Logarithmic Functions

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In summary, the conversation includes two questions: finding the 2nd derivative at a point and evaluating the first derivative at a given value. The first question involves substituting the given point into the 2nd derivative equation, while the second question can be solved using either the quotient rule, chain rule, and product rule, or through first principles. The speaker also suggests a trick for simplifying the second function by using negative exponents.
  • #1
TommyLF
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I need a little help on the following problems:

Find the 2nd derivative at the indicated point:
y= x(ln x) At the point (1,0)

I got to:

y''= [tex]
\frac{1}{x}
[/tex]

So do I just sub in 1 for x? And then the answer would be 1?

Also, I need help on the following problem:

Evaluate the first derivative at the given value X: x=0

[4^(3x)(x^3 - x + 1)^(1/5)] (And then all of that raised to the -2)
-------------------------
(x^2 + x + 1)^(4)
 
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  • #2
TommyLF said:
I need a little help on the following problems:

Find the 2nd derivative at the indicated point:
y= x(ln x) At the point (1,0)

I got to:

y''= [tex]
\frac{1}{x}
[/tex]

So do I just sub in 1 for x? And then the answer would be 1?

Also, I need help on the following problem:

Evaluate the first derivative at the given value X: x=0

[4^(3x)(x^3 - x + 1)^(1/5)] (And then all of that raised to the -2)
-------------------------
(x^2 + x + 1)^(4)

Yeah, basically put 1 in the equation for the first question.

For the second one, I see no trick. I guess you have to painfully use the Quotient Rule, Chain Rule and Product in the right order. Have fun doing that. :S

Actually, I'd probably do it using first principles. It seems like the easier way out. (First principles is finding the derivative using the limit definition.)
 
  • #3
JasonRox, you have a wicked sense of humor! Anyone trying to find the derivative of that second function using the limit of the difference quotient is guarenteed to go insane!

TommyLF, I would recommend going ahead and incorporating that -2 exponent into the numerator and, because that is a negative exponent, writing the whole thing as a product of terms with negative exponents:
[tex]\frac{[4^{3x}(x^3- x+ 1)^{1/5}]^{-2}}{(x^2+ x+ 1)^4}= 4^{-6x}(x^2-x+1)^{-2/5}(x^2+ x+ 1)^{-4}[/tex]
and use the product rule and chain rule.
 

Related to Derivative Problems with Exponential and Logarithmic Functions

What is a derivative?

A derivative is a mathematical concept that represents the rate of change of one variable with respect to another. In other words, it measures how much one quantity changes in response to a change in another quantity.

Why are derivatives important?

Derivatives are important in many areas of science and mathematics, including physics, economics, and engineering. They allow us to model and analyze complex systems and make predictions about how they will behave in the future.

How do you find the derivative of a function?

The derivative of a function can be found using various techniques, including the power rule, product rule, quotient rule, and chain rule. These techniques involve manipulating the function algebraically and using the properties of derivatives to simplify the expression.

What are some real-life applications of derivatives?

Derivatives have numerous real-life applications, such as in physics to calculate velocity and acceleration, in economics to determine marginal cost and revenue, and in engineering to optimize designs and control systems.

What are some common mistakes when working with derivatives?

Some common mistakes when working with derivatives include forgetting to apply the chain rule, making algebraic errors, and not simplifying the final expression. It is important to double-check your work and practice regularly to avoid these mistakes.

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