Originally Posted by
earboth Do you really mean:
$\displaystyle f(x) = \left(\sin(x)\right)^{4x}$ ...... ??
If so re-write f to:
$\displaystyle f(x) = e^{4x \cdot \ln(\sin(x))}$ ...... and use the chain rule to calculate the derivative:
$\displaystyle f'(x) = \frac{e^{4x \cdot \ln(\sin(x))}}{\sin(x)} \cdot \left(4 \sin(x) \cdot \ln(\sin(x)) + 4x \cdot \cos(x) \right)$
Now plug in $\displaystyle x = \frac{\pi}{2}$ to calculate the slope of the tangent.
Since $\displaystyle \ln\left(\sin\left(\frac{\pi}2 \right) \right) = 0$ and $\displaystyle \cos \left(\frac{\pi}2 \right) = 0$ too you'll get:
$\displaystyle f'\left(\frac{\pi}2 \right) = 0$ and therefore the equation of the tangent is
$\displaystyle y = 0$