Finding g'(x) in a(x)=x(18-x^2)^1/2

  • Thread starter Thread starter Shambles
  • Start date Start date
Shambles
Messages
14
Reaction score
0

Homework Statement


Find g'(x)
a(x)=x(18-x^2)^1/2

Homework Equations


Answer stated as:
a'(x)=(18-x^2)^1/2 - x^2/(18-x^2)^1/2


The Attempt at a Solution



Having trouble with this solution. The chain rule states that f(x)g(x) = f'(x)g(x)+g'(x)f(x) so the first term in the solution is obviously (18-x^2)^1/2.

Where I run into problems is the second term with the numerator being x^2 when I thought it should only be x. Simplifying the problem into f(x)=(18-x^2)^1/2 the chain rule states it the derivative should be 1/2(18-x^2)(0-2x)=-x/(18-x^2)^1/2. Am I wrong or is the stated solution wrong?
 
Physics news on Phys.org
Shambles said:
Having trouble with this solution. The chain rule states that f(x)g(x) = f'(x)g(x)+g'(x)f(x) so the first term in the solution is obviously (18-x^2)^1/2.

Where I run into problems is the second term with the numerator being x^2 when I thought it should only be x.
You forgot about f(x) = x, which gives you the second x.
 
Ugh, that is so... dirty. Thanks, that was pretty simple.
 
Shambles said:

Homework Statement


Find g'(x)
a(x)=x(18-x^2)^1/2

Homework Equations


Answer stated as:
a'(x)=(18-x^2)^1/2 - x^2/(18-x^2)^1/2


The Attempt at a Solution



Having trouble with this solution. The chain rule states that f(x)g(x) = f'(x)g(x)+g'(x)f(x) so the first term in the solution is obviously (18-x^2)^1/2.
That's not the chain rule; it's the product rule. The chain rule is used to find the derivative of a composite function, f(g(x)).
Shambles said:
Where I run into problems is the second term with the numerator being x^2 when I thought it should only be x. Simplifying the problem into f(x)=(18-x^2)^1/2 the chain rule states it the derivative should be 1/2(18-x^2)(0-2x)=-x/(18-x^2)^1/2. Am I wrong or is the stated solution wrong?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
View full post »

Similar threads

Why Is the Chain Rule Not Used in Differentiating h(x) = 3f(x) + 8g(x)?
Replies
2
Views
2K
Relationship between autonomous system and related single equation
Replies
3
Views
2K
Why Does the Multidimensional Chain Rule Lead to Euler's Theorem Proof?
Replies
2
Views
1K
Variation of parameter VS Undetermined Coefficients
Replies
7
Views
2K
Find g(x)/h(y) for a given F(x,y)
Replies
3
Views
1K
Prove that the integral is equal to ##\pi^2/8##
3
Replies
105
Views
4K
Taylor Expansion for very small and very big arguments
Replies
1
Views
2K
Proving stability of linear system equations
Replies
2
Views
1K
How do you find the missing maximum of g(x)?
Replies
28
Views
2K
Given unit impulse response, obtain differential equation
Replies
7
Views
1K

Hot Threads

Recent Insights

Back
Top