Detail solution
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Let .
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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
[src]
tan(t) / 2 \
a *\1 + tan (t)/*log(a)
$$a^{\tan{\left(t \right)}} \left(\tan^{2}{\left(t \right)} + 1\right) \log{\left(a \right)}$$
The second derivative
[src]
tan(t) / 2 \ / / 2 \ \
a *\1 + tan (t)/*\2*tan(t) + \1 + tan (t)/*log(a)/*log(a)
$$a^{\tan{\left(t \right)}} \left(\left(\tan^{2}{\left(t \right)} + 1\right) \log{\left(a \right)} + 2 \tan{\left(t \right)}\right) \left(\tan^{2}{\left(t \right)} + 1\right) \log{\left(a \right)}$$
The third derivative
[src]
/ 2 \
tan(t) / 2 \ | 2 / 2 \ 2 / 2 \ |
a *\1 + tan (t)/*\2 + 6*tan (t) + \1 + tan (t)/ *log (a) + 6*\1 + tan (t)/*log(a)*tan(t)/*log(a)
$$a^{\tan{\left(t \right)}} \left(\tan^{2}{\left(t \right)} + 1\right) \left(\left(\tan^{2}{\left(t \right)} + 1\right)^{2} \log{\left(a \right)}^{2} + 6 \left(\tan^{2}{\left(t \right)} + 1\right) \log{\left(a \right)} \tan{\left(t \right)} + 6 \tan^{2}{\left(t \right)} + 2\right) \log{\left(a \right)}$$