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Math Help - Differentiation help

  1. #1
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    Differentiation help

    I need help with these two questions that for some reason I keep getting wrong. I am suppose to differentiate with respect to x

    sin((x^2+3)^4)

    tan^5(x)sec^3(X)

    For the last one I know the product rule needs to be used.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by NYC08 View Post
    I need help with these two questions that for some reason I keep getting wrong. I am suppose to differentiate with respect to x

    sin((x^2+3)^4)

    tan^5(x)sec^3(X)

    For the last one I know the product rule needs to be used.
    You need to apply chain rule to the first one:

    \begin{aligned}y=\sin\left[\left(x^2+3\right)^4\right]\implies \frac{\,dy}{\,dx} & = \cos\left[\left(x^2+3\right)^4\right]\cdot\frac{\,d}{\,dx}\left[\left(x^2+3\right)^4\right]\\ & = \cos\left[\left(x^2+3\right)^4\right]\cdot4\left(x^2+3\right)^3\cdot\frac{\,d}{\,dx}\le  ft[x^2+3\right]\\ &= \dots\end{aligned}

    Can you finish this one?

    For the second one, apply product and chain rule.

    \begin{aligned}<br />
y=\tan^5x\sec^3x\implies \frac{\,dy}{\,dx} & = 5\tan^4x\cdot\frac{\,d}{\,dx}\left(\tan x\right)\cdot\sec^3x+\tan^5x\cdot3\sec^2x\cdot\fra  c{\,d}{\,dx}(\sec x)\\<br />
&=\dots\end{aligned}

    Can you finish this one?

    Does this make sense?
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  3. #3
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    For the first one I get

    8x(x^2+3)^3 cos((x^2+3)^4)

    using the chain rule (twice)

    and for the second notice

    tan^5(x)sec^3(x) = sin^5(x)/cos^8(x)

    so differentiating using the quotient rule, I get

    (5sin^4(x)cos^9(x) + 8cos^7(x)sin^6(x))/cos^16(x)

    which will cancel down a bit.
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