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Thread: another differentiation of a function (bit harder!)

  1. #1
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    another differentiation of a function (bit harder!)

    v = 3 sin (u) - 10tan (u) + 101n (u) + 2e^u

    the answer i got was

    3cos (u) - 10/cos^2 (u) + 101/u + 2ue^u-1

    maybe not right stuck with what to do on the -10tan and the last part
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by JontyP View Post
    v = 3 sin (u) - 10tan (u) + 101n (u) + 2e^u

    the answer i got was

    3cos (u) - 10/cos^2 (u) + 101/u + 2ue^u-1

    maybe not right stuck with what to do on the -10tan and the last part
    the derivative of $\displaystyle e^x$ is $\displaystyle e^x$

    the derivative of a function of the form $\displaystyle a^x$, where $\displaystyle a>0$ is a constant and $\displaystyle x$ is our variable, is $\displaystyle a^x \ln a$

    what is 101n (u) ?

    what is the n? are there any powers here? (you should use ^ to indicate powers, so we know what you're talking about)
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  3. #3
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    i did use ^ on the 2e ^ u is supposed to read 2e to the power of u

    the n is just in the question it reads +101n (u) but im not 100% sure what to do with that

    the e^u yeh that should be 2e^u after readin my notes again.

    3cos (u) - 10/cos^2 (u) + 101/u + 2e^u

    big question mark on the 101/u part however
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by JontyP View Post
    i did use ^ on the 2e ^ u is supposed to read 2e to the power of u

    the n is just in the question it reads +101n (u) but im not 100% sure what to do with that

    the e^u yeh that should be 2e^u after readin my notes again.

    3cos (u) - 10/cos^2 (u) + 101/u + 2e^u

    big question mark on the 101/u part however
    for 101n (u), are we to assume it looks like $\displaystyle 101nu$? in that case, it's just using the power rule...

    wait did you mean $\displaystyle \ln u$, as in the natural log of u? it is not a 1, it is a lower case L!

    so, if you have $\displaystyle 10 \ln u$ the derivative is $\displaystyle \frac {10}u$

    and as i said, write $\displaystyle \sec^2 x$ instead of $\displaystyle \frac 1{\cos^2 x}$, it looks nicer

    the rest are ok
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