Find and simplify the following derivatives: $\displaystyle f(x) = cos^3 (3x^2 - 2) $ $\displaystyle f(x) = sec^2 (1 - \sqrt{4x}) $ $\displaystyle f(x) = tan^3 (\sqrt{1-x^3}) $ thanks
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all of these are derivatives involving the use the chain rule ... let's see how you do at least one of them.
i know they are the chain rule which is $\displaystyle g'(h(x)) \cdot h'(x) $ but I dont know how to do these chain rules with cos, sec and tan to exponents of more then 1
$\displaystyle f(x) = [\sec(1 - \sqrt{4x})]^2$ $\displaystyle f'(x) = 2\sec(1 - \sqrt{4x}) \cdot \sec(1 - \sqrt{4x}) \cdot \tan(1 - \sqrt{4x}) \cdot \left(-\frac{1}{\sqrt{x}}\right)$ do the cosine one.
Originally Posted by skeeter $\displaystyle f(x) = [\sec(1 - \sqrt{4x})]^2$ $\displaystyle f'(x) = 2\sec(1 - \sqrt{4x}) \cdot \sec(1 - \sqrt{4x}) \cdot \tan(1 - \sqrt{4x}) \cdot \left(-\frac{1}{\sqrt{x}}\right)$ do the cosine one. wouldnt this one be: $\displaystyle f'(x) = 2sec(1 - \sqrt{4x}) \cdot sec(1 - \sqrt{4x}) \cdot tan(1 - \sqrt{4x}) \cdot (\frac{-1}{2\sqrt{4x}}) $ ???
$\displaystyle \frac{d}{dx} (-\sqrt{4x}) = $ $\displaystyle \frac{d}{dx} (-2\sqrt{x}) = $ $\displaystyle -2 \cdot \frac{d}{dx} (\sqrt{x}) = -2 \cdot \frac{1}{2\sqrt{x}} = -\frac{1}{\sqrt{x}}$
so the cos one would be $\displaystyle f'(x) = 3[cos(3x^2-2)]^2 \cdot [cos(3x^2-2)] \cdot [-sin(3x^2-2)] \cdot (6x) $
Originally Posted by iz1hp so the cos one would be $\displaystyle f'(x) = 3[cos(3x^2-2)]^2 \cdot [cos(3x^2-2)] \cdot [-sin(3x^2-2)] \cdot (6x) $ almost ... $\displaystyle f'(x) = 3[cos(3x^2-2)]^2 \cdot [-sin(3x^2-2)] \cdot (6x)$
so the tan one is... $\displaystyle f(x) = tan^3(\sqrt{1-x^3}) $ $\displaystyle f(x) = [tan(\sqrt{1-x^3})]^3 $ $\displaystyle f'(x) = 3[tan(\sqrt{1-x^3})]^2 \cdot [sec^2(\sqrt{1-x^3})] \cdot (\frac{1}{2\sqrt{1-x^3}}) \cdot (-3x^2) $ ??
... and you said you didn't know how. good job. now clean up the algebra on all of them.
ahhah thanks for explaining it to me haha :P if you can help me with one more, ill be set :P $\displaystyle f(x) = cos(sin(cos x)) $ so far ive gotten to: $\displaystyle f'(x) = -sin(sin(cos x)) \cdot cos(cos x) \cdot -sin x $ and now im stuck
Originally Posted by iz1hp ahhah thanks for explaining it to me haha :P if you can help me with one more, ill be set :P $\displaystyle f(x) = cos(sin(cos x)) $ so far ive gotten to: $\displaystyle f'(x) = -sin(sin(cos x)) \cdot cos(cos x) \cdot -sin x $ and now im stuck Hey iz1hp. How about starting a new thread for this new question? It's a habit I'd like you and all users to get into. I'm going to close this thread but please make a new one! Jameson
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