Derivative of f(x) = x(20−x)/36 and composite function g(x) = e^f(x)

[ghostbuster25]

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)=e^x(20-x)/36

g'(x)=f'(x)*e^f(x)

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

is this correct?

is this correct

 

[Mark44]

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) = ...

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

Above - Now you have taken the derivative

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

Above - now you have an error. 20 - 2x != 2(10 - 2x).

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

g'(x)=f'(x)*e^f(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.

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

is this correct?

is this correct