Is This Derivation of the Function Correct?

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ok just want a check of my work before i send it :)

differntiate the function;
f(x)=\frac{x(20-x)}{36}

so,
f'(x)=\frac{(20x-x^2)}{36}

f'(x)=\frac{(20-2x)}{36}

f'(x)=\frac{(10-2x)}{18}

Then...use the composite rule and your answer above to differntiate the function
g(x)=ex(20-x)/36

g'(x)=f'(x)*ef(x)

so,
g'(x)=\frac{10-2x)}{18}*g(x)=ex(20-x)/36

is this correct?

is this correct
 
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ghostbuster25 said:
ok just want a check of my work before i send it :)

differntiate the function;
f(x)=\frac{x(20-x)}{36}

so,
f'(x)=\frac{(20x-x^2)}{36}
Above - This is not f'(x). All you have done is expand the numerator on the right. This should be f(x) = ...
ghostbuster25 said:
f'(x)=\frac{(20-2x)}{36}
Above - Now you have taken the derivative
ghostbuster25 said:
f'(x)=\frac{(10-2x)}{18}
Above - now you have an error. 20 - 2x != 2(10 - 2x).
ghostbuster25 said:
Then...use the composite rule and your answer above to differntiate the function
g(x)=ex(20-x)/36

g'(x)=f'(x)*ef(x)
Where f(x) = (1/36)(x(20 - x)). The derivative below is incorrect. To fix it, replace f'(x) and f(x) where they appear.
ghostbuster25 said:
so,
g'(x)=\frac{10-2x)}{18}*g(x)=ex(20-x)/36

is this correct?

is this correct
 
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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