# D/dx 6x^(3/2)tan(x) = 3x^(1/2)(2xsec^2(x)+3tan(x))? ?

• zketrouble
In summary, the conversation is about differentiating the expression 6x^(3/2)tan(x) using the product rule. The person asking the question initially has trouble factoring their answer, but after some help from others, realizes their mistake.
zketrouble
d/dx 6x^(3/2)tan(x) = 3x^(1/2)(2xsec^2(x)+3tan(x))?!?

Hi all,

I've been officially bored to death with my College Algebra class so I just started teaching myself Calculus about two months ago. I found a list of problems using the product rule http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/productruledirectory/ProductRule.html#PROBLEM 4". Problem 4 is as follows:

Differentiate 6x^(3/2)tan(x)

Doesn't look too hard, I start by remembering that (fg)' = f'g +fg'.

d/dx f(x) = d/dx[6x^(3/2)]tan(x) + 6x^(3/2)d/dx[tan(x)]
d/dx f(x) = 9x^(1/2)tan(x) + 6x^(3/2)sec^2(x)

So I clicked the link on that site to view the detailed solution, which says the answer is 3x^(1/2)(2xsec^2(x)+3tan(x)). I hate to make the assumption that it is incorrect being a calculus noob and all, but it appears that 3x^(1/2)(2xsec^2(x)+3tan(x)) is incorrectly factored, because 3x^(1/2) times 3tan(x) does not equal 6x^(3/4)tan(x). The websites solution says that my answer is the step before they arrive at this solution which I assume to be false, am I missing something here?

Thanks!

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3x^(1/2)(2xsec^2(x)+3tan(x)) is correctly factored. Work it out: 3^(1/2)*3=9^(1/2) and 2x*3x^(1/2)=6x^(3/2) I'm not sure where you're getting 6x^(3/4)tan(x). But other than that you did everything correctly.

zketrouble said:
Hi all,

I've been officially bored to death with my College Algebra class so I just started teaching myself Calculus about two months ago. I found a list of problems using the product rule http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/productruledirectory/ProductRule.html#PROBLEM 4". Problem 4 is as follows:

Differentiate 6x^(3/2)tan(x)

Doesn't look too hard, I start by remembering that (fg)' = f'g +fg'.

d/dx f(x) = d/dx[6x^(3/2)]tan(x) + 6x^(3/2)d/dx[tan(x)]
d/dx f(x) = 9x^(1/2)tan(x) + 6x^(3/2)sec^2(x)

So I clicked the link on that site to view the detailed solution, which says the answer is 3x^(1/2)(2xsec^2(x)+3tan(x)). I hate to make the assumption that it is incorrect being a calculus noob and all, but it appears that 3x^(1/2)(2xsec^2(x)+3tan(x)) is incorrectly factored, because 3x^(1/2) times 3tan(x) does not equal 6x^(3/4)tan(x). The websites solution says that my answer is the step before they arrive at this solution which I assume to be false, am I missing something here?Thanks!

$$3x^{\frac{1}{2}} \left( 2x\sec ^ 2 (x) + 3 \tan x \right) = 3x^{\frac{1}{2}} 2x\sec ^ 2 (x) + 3 \times 3x^{\frac{1}{2}} \tan x$$
$$= 6x^{\frac{1}{2} + 1}\sec ^ 2 (x) + 9x^{\frac{1}{2}} \tan x = 6x^{\frac{3}{2}}\sec ^ 2 (x) + 9x^{\frac{1}{2}} \tan x$$

Looks good to me. :) And it looks exactly as what you've got, too.

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armolinasf said:
3x^(1/2)(2xsec^2(x)+3tan(x)) is correctly factored. Work it out: 3^(1/2)*3=9^(1/2) and 2x*3x^(1/2)=6x^(3/2) I'm not sure where you're getting 6x^(3/4)tan(x). But other than that you did everything correctly.

Ok silly mistake on my part...just like nearly all of my mistakes. Thanks.

## 1. What is the meaning of "d/dx" in this equation?

"d/dx" represents the derivative of the given function with respect to x. It is a mathematical operation that calculates the rate of change of the function at a specific point.

## 2. How do you solve this equation?

To solve this equation, you can use the power rule and the product rule of differentiation. First, take the derivative of 6x^(3/2) which is 9x^(1/2). Then, take the derivative of tan(x) which is sec^2(x) and multiply it by 6x^(3/2). Finally, use the product rule to differentiate 6x^(3/2)tan(x) and simplify the equation to its final form.

## 3. What is the purpose of finding the derivative of this function?

The derivative of a function tells us the slope of the function at any given point. In this case, the derivative of 6x^(3/2)tan(x) gives us the slope of the function at any x-value. It can also be used to find the rate of change of a variable in a given system.

## 4. Can this equation be used in real-life applications?

Yes, this equation can be used in various fields such as physics, engineering, and economics. For example, in physics, the derivative can be used to find the velocity and acceleration of an object in motion. In economics, it can be used to calculate the marginal cost and revenue of a product.

## 5. How can I check if my solution to this equation is correct?

You can use a graphing calculator or software to graph both sides of the equation and see if they intersect at the same point. You can also take the derivative of the final solution and see if it matches the original equation.

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