# Thread: Check derivtives for correctness

1. ## Check derivtives for correctness

I have not done differential calculus for awhile. I reviewed my calc book, but I am still having some trouble. Could someone please check these derivatives?

$\displaystyle f(x) = e^x + 2^-^x + 2*cos(x)-6$
$\displaystyle f'(x) = e^x + ln(2)*2^-^x - 2*sin(x)-0$
$\displaystyle f'(x) = e^x + 2^-^x * ln(2) - 2*sin(x)$

$\displaystyle g(x) = 2xcos(2x)-(x-2)^2$
$\displaystyle g'(x) = 2x (d/dx* cos(2*x))+cos(2*x)*(d/dx* 2x) - 2(x-2)$
$\displaystyle g'(x) = -2xsin(2x)(2)+cos(2x)*2-2(x-2)$
$\displaystyle g'(x) = -4xsin(2x) + 2cos(2x) - 2(x-2)$

$\displaystyle h(x) = e^x - 3x^2$
$\displaystyle h'(x) = e^x - 3*2*x$
$\displaystyle h'(x) = e^x - 6*x$

2. Originally Posted by sedemihcra
I have not done differential calculus for awhile. I reviewed my calc book, but I am still having some trouble. Could someone please check these derivatives?

$\displaystyle f(x) = e^x + 2^-^x + 2*cos(x)-6$
$\displaystyle f'(x) = e^x + ln(2)*2^-^x - 2*sin(x)-0$
$\displaystyle f'(x) = e^x + 2^-^x * ln(2) - 2*sin(x)$

$\displaystyle g(x) = 2xcos(2x)-(x-2)^2$
$\displaystyle g'(x) = 2x (d/dx* cos(2*x))+cos(2*x)*(d/dx* 2x) - 2(x-2)$
$\displaystyle g'(x) = -2xsin(2x)(2)+cos(2x)*2-2(x-2)$
$\displaystyle g'(x) = -4xsin(2x) + 2cos(2x) - 2(x-2)$

$\displaystyle h(x) = e^x - 3x^2$
$\displaystyle h'(x) = e^x - 3*2*x$
$\displaystyle h'(x) = e^x - 6*x$
Correct.

3. The first one:

(2^{-x} )'=-ln(2)2^{-x}

The third one is correct.

4. [QUOTE=sedemihcra;606004]I have not done differential calculus for awhile. I reviewed my calc book, but I am still having some trouble. Could someone please check these derivatives?

$\displaystyle f(x) = e^x + 2^-^x + 2*cos(x)-6$
$\displaystyle f'(x) = e^x + ln(2)*2^-^x - 2*sin(x)-0$
$\displaystyle f'(x) = e^x + 2^-^x * ln(2) - 2*sin(x)$
The second term should be $\displaystyle -2^{-x} ln(2)$ because of the - in the numerator.
By the way, the it is simpler to get $\displaystyle 2^{-x}$ using "braces": "2^{-x}" rather than "2^-^x".

$\displaystyle g(x) = 2xcos(2x)-(x-2)^2$
$\displaystyle g'(x) = 2x (d/dx* cos(2*x))+cos(2*x)*(d/dx* 2x) - 2(x-2)$
$\displaystyle g'(x) = -2xsin(2x)(2)+cos(2x)*2-2(x-2)$
$\displaystyle g'(x) = -4xsin(2x) + 2cos(2x) - 2(x-2)$
Good!

$\displaystyle h(x) = e^x - 3x^2$
$\displaystyle h'(x) = e^x - 3*2*x$
$\displaystyle h'(x) = e^x - 6*x$
Good.

5. Originally Posted by Also sprach Zarathustra
The first one:

(2^{-x} )'=-ln(2)2^{-x}

The third one is correct.
Good eye on the negative sign.

6. Thanks for all the help. Also thanks for getting back so quick.