Calculating Derivatives Using the Power Rule and Chain Rule

[ciubba]

I am having difficulty calculating the following derivative $${ \frac{2x^2-1}{(3x^4+2)^2}}$$

Could someone demonstrate the first step algebraically? Assuming c is the exponent on the variable expression, n is the numerator and d is the denominator, I tried:

$$c\frac{n(x)}{d(x)}\frac{n'(x)*d(x)-d'(x)*n(x)}{d(x)^2}$$

Which gives me
$$2\frac{2x^2-1}{3x^4+2}\frac{[4x(3x^4+2)]-[(12x^3)(2x^2-1)]}{(3x^4+2)^2}$$

Which simplifies to
$$\frac{-48 x^7+72 x^5+8 x^3-16 x}{(3x^4+2)^3}$$

However, the book lists the answer as being $$\frac{-36x^5+24x^3+8x}{(3x^4+2)^2}$$

 

[Char. Limit]

I'm not quite sure what rule you're trying to use, but if it's the quotient rule, then you've got it written down wrong. The correct way is:

$$\frac{d}{dx} \frac{n(x)}{d(x)} = \frac{n'(x) d(x) - d'(x) n(x)}{d(x)^2}$$

Since you've already worked the latter expression out, it should be easy to finish for you.

 

[NathanaelNolk]

Are you sure that : $$\frac{d}{dx} (3x^4+2)^2=12x^3$$ ?

 

[ciubba]

I'm not quite sure what rule you're trying to use, but if it's the quotient rule, then you've got it written down wrong. The correct way is:

$$\frac{d}{dx} \frac{n(x)}{d(x)} = \frac{n'(x) d(x) - d'(x) n(x)}{d(x)^2}$$

Since you've already worked the latter expression out, it should be easy to finish for you.

My book lists a rule called the "extended power rule," which goes as follows:

"Suppose g(x) is a differentiable function of x. Then, for any real number k,

$$\frac {d}{dx}[g(x)]^k=k[g(x)]^{k-1}*\frac{d}{dx}[g(x)]$$

Here's a link to the text:

http://www.scribd.com/doc/13142109/1-7-the-Chain-Rule

I could easily solve the problem by expanding the binomial expression (3x^4+2)^2 and then using the standard product rule, but I need to know how to use the extended product rule as one of the sample question is raised to the power of 7, and there is no way that I am going to expand that. If you can offer me an alternative method to dealing with the derivatives of high order expressions, I would accept that as well.

 

[Char. Limit]

The extended power rule isn't exactly relevant here. And you got that particular part correct. The problem was that the quotient rule, for whatever reason, was done incorrectly. I'm not sure where you got a c or the first part of that product.

 

[ciubba]

The extended power rule isn't exactly relevant here. And you got that particular part correct. The problem was that the quotient rule, for whatever reason, was done incorrectly. I'm not sure where you got a c or the first part of that product.

The book says to use the extended power rule in addition to the quotient rule to solve this particular problem, so it has to be relevant >_>

According to the extended power rule, I take the exponent off the expression, k (i accidentally put c), and multiply it by g(x). The text has a step-by-step example of how to use the extended power rule in conjunction with another quotient problem, $$\sqrt[4]{\frac{x+3}{x-2}}$$ in which they use the setup $$k\frac{n(x)}{d(x)}\frac{n'(x)*d(x)-d'(x)*n(x)}{d(x)^2}$$, but that form doesn't appear to work here.

 

[NathanaelNolk]

I could easily solve the problem by expanding the binomial expression (3x^4+2)^2 and then using the standard product rule, but I need to know how to use the extended product rule as one of the sample question is raised to the power of 7, and there is no way that I am going to expand that. If you can offer me an alternative method to dealing with the derivatives of high order expressions, I would accept that as well.

You don't need to expend these high order derivatives, just use the chain rule. As a reminder $$(3x^4+2)^2$$ can be seen as a function of this type : $$f(g(x))\ \mbox{where}\ g(x)=3x^4+2\ \mbox{and}\ f(x) = x^2\\ \mbox {Now consider that the x in}\ x^2\ \mbox{actually is your function g(x), that is, f(x) is applied to g(x), the x in brackets become g(x).}\\ \mbox{You then have your function h(x) = f(g(x)), which is} (3x^4+2)^2\\ \mbox{You can now differentiate h(x), and as you can see, it's simply the derivative of f(g(x)).}\\ \mbox{Use the chain rule:} \frac{d}{dx}f(g(x)) = f'(g(x)) * g'(x)\ \mbox{and you've got the derivative.}\\ \mbox{You can now differentiate polynomials, for instance :} \frac{d}{dx}(4x^5+3)^9 = 9*(4x^5+3)^8 * 20x^4\\ \mbox{In general :}\ \frac{d}{dx}(P(x))^n = n(P(x))^{n-1} * P'(x)$$

Hope this helps!

 

[ciubba]

you don't need to expend these high order derivatives, just use the chain rule. As a reminder $$(3x^4+2)^2$$ can be seen as a function of this type : $$f(g(x))\ \mbox{where}\ g(x)=3x^4+2\ \mbox{and}\ f(x) = x^2\\ \mbox {now consider that the x in}\ x^2\ \mbox{actually is your function g(x), that is, f(x) is applied to g(x), the x in brackets become g(x).}\\ \mbox{you then have your function h(x) = f(g(x)), which is} (3x^4+2)^2\\ \mbox{you can now differentiate h(x), and as you can see, it's simply the derivative of f(g(x)).}\\ \mbox{use the chain rule:} \frac{d}{dx}f(g(x)) = f'(g(x)) * g'(x)\ \mbox{and you've got the derivative.}\\ \mbox{you can now differentiate polynomials, for instance :} \frac{d}{dx}(4x^5+3)^9 = 9*(4x^5+3)^8 * 20x^4\\ \mbox{in general :}\ \frac{d}{dx}(p(x))^n = n(p(x))^{n-1} * p'(x)$$

hope this helps!

thank you!