# Check derivtives for correctness

• Jan 16th 2011, 11:58 AM
sedemihcra
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\$
• Jan 16th 2011, 12:00 PM
dwsmith
Quote:

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.
• Jan 16th 2011, 12:04 PM
Also sprach Zarathustra
The first one:

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

The third one is correct.
• Jan 16th 2011, 12:05 PM
HallsofIvy
[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".

Quote:

\$\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!

Quote:

\$\displaystyle h(x) = e^x - 3x^2\$
\$\displaystyle h'(x) = e^x - 3*2*x\$
\$\displaystyle h'(x) = e^x - 6*x\$
Good.
• Jan 16th 2011, 12:05 PM
dwsmith
Quote:

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.
• Jan 16th 2011, 12:09 PM
sedemihcra
Thanks for all the help. Also thanks for getting back so quick.