Check derivtives for correctness

**sedemihcra** · January 16th 2011, 11:58 AM

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?

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

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

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

**dwsmith** · January 16th 2011, 12:00 PM

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?

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

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

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

Correct.

**Also sprach Zarathustra** · January 16th 2011, 12:04 PM

The first one:

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

The third one is correct.

**HallsofIvy** · January 16th 2011, 12:05 PM

[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?

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

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

Good!

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

Good.

**dwsmith** · January 16th 2011, 12:05 PM

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.

**sedemihcra** · January 16th 2011, 12:09 PM

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

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