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Math Help - 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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    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 e^x is e^x

    the derivative of a function of the form a^x, where a>0 is a constant and x is our variable, is 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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    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 101nu? in that case, it's just using the power rule...

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

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

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

    the rest are ok
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