Detail solution
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Apply the quotient rule, which is:
and .
To find :
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Rewrite the function to be differentiated:
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Apply the quotient rule, which is:
and .
To find :
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The derivative of sine is cosine:
To find :
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The derivative of cosine is negative sine:
Now plug in to the quotient rule:
To find :
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Now plug in to the quotient rule:
Now simplify:
The answer is:
The first derivative
[src]
-x / 2 \ -x
2 *\1 + tan (x)/ - 2 *log(2)*tan(x)
$$2^{- x} \left(\tan^{2}{\left(x \right)} + 1\right) - 2^{- x} \log{\left(2 \right)} \tan{\left(x \right)}$$
The second derivative
[src]
-x / 2 / 2 \ / 2 \ \
2 *\log (2)*tan(x) - 2*\1 + tan (x)/*log(2) + 2*\1 + tan (x)/*tan(x)/
$$2^{- x} \left(2 \left(\tan^{2}{\left(x \right)} + 1\right) \tan{\left(x \right)} - 2 \left(\tan^{2}{\left(x \right)} + 1\right) \log{\left(2 \right)} + \log{\left(2 \right)}^{2} \tan{\left(x \right)}\right)$$
The third derivative
[src]
-x / 3 / 2 \ / 2 \ 2 / 2 \ / 2 \ \
2 *\- log (2)*tan(x) + 2*\1 + tan (x)/*\1 + 3*tan (x)/ + 3*log (2)*\1 + tan (x)/ - 6*\1 + tan (x)/*log(2)*tan(x)/
$$2^{- x} \left(2 \left(\tan^{2}{\left(x \right)} + 1\right) \left(3 \tan^{2}{\left(x \right)} + 1\right) - 6 \left(\tan^{2}{\left(x \right)} + 1\right) \log{\left(2 \right)} \tan{\left(x \right)} + 3 \left(\tan^{2}{\left(x \right)} + 1\right) \log{\left(2 \right)}^{2} - \log{\left(2 \right)}^{3} \tan{\left(x \right)}\right)$$