Optimizing the Product Rule: Common Error Help for (5x-2)^3 / (2x+5)^4

fr33pl4gu3
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y = ( 5 x-2 ) 3 / ( 2 x+5 ) 4

I do it this way:

f(x) = (5x-2)3
g(x) = (2x+5)-4
f'(x) = 3(5x-2)2
g'(x) = -4(2x+5)-5

By using the product rule:

[3(5x-2)2](2x+5)-4 + (5x-2)3[-4(2x+5)-5]
3(5x-2)2(2x+5)-4 -4(2x+5)-5(5x-2)3

What's wrong with this answer?
 
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Thread moved to Homework Help

f'(x) = 3(5x-2)2
g'(x) = -4(2x+5)-5

I do not believe you have differentiated correctly in these steps.
 


Is it this way??

f'(x) = 15(5x-2)2
g'(x) = -8(2x+5)-5
 
Last edited:


Yes, got it. Thanks for pointing out the error.
 

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...
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