Detail solution
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Let .
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The derivative of cosine is negative sine:
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Then, apply the chain rule. Multiply by :
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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:
The result of the chain rule is:
Now simplify:
The answer is:
The first derivative
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-\1 + tan (x)/*sin(tan(x))
$$- \left(\tan^{2}{\left(x \right)} + 1\right) \sin{\left(\tan{\left(x \right)} \right)}$$
The second derivative
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/ 2 \ // 2 \ \
-\1 + tan (x)/*\\1 + tan (x)/*cos(tan(x)) + 2*sin(tan(x))*tan(x)/
$$- \left(\left(\tan^{2}{\left(x \right)} + 1\right) \cos{\left(\tan{\left(x \right)} \right)} + 2 \sin{\left(\tan{\left(x \right)} \right)} \tan{\left(x \right)}\right) \left(\tan^{2}{\left(x \right)} + 1\right)$$
The third derivative
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/ 2 \
/ 2 \ |/ 2 \ 2 / 2 \ / 2 \ |
\1 + tan (x)/*\\1 + tan (x)/ *sin(tan(x)) - 4*tan (x)*sin(tan(x)) - 2*\1 + tan (x)/*sin(tan(x)) - 6*\1 + tan (x)/*cos(tan(x))*tan(x)/
$$\left(\tan^{2}{\left(x \right)} + 1\right) \left(\left(\tan^{2}{\left(x \right)} + 1\right)^{2} \sin{\left(\tan{\left(x \right)} \right)} - 2 \left(\tan^{2}{\left(x \right)} + 1\right) \sin{\left(\tan{\left(x \right)} \right)} - 6 \left(\tan^{2}{\left(x \right)} + 1\right) \cos{\left(\tan{\left(x \right)} \right)} \tan{\left(x \right)} - 4 \sin{\left(\tan{\left(x \right)} \right)} \tan^{2}{\left(x \right)}\right)$$