find the equation of the tangent line to y=(sin x)^4x at the point (pi/2, 1)

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- Feb 10th 2008, 11:24 PMBust2000tangent line
find the equation of the tangent line to y=(sin x)^4x at the point (pi/2, 1)

- Feb 10th 2008, 11:55 PMJhevon
a line is of the form $\displaystyle y = mx + b$

we will find the line by the point-slope form. that is, $\displaystyle y - y_1 = m(x - x_1)$, where $\displaystyle m$ is the slope (given by the derivative evaluated at $\displaystyle (x_1,y_1)$) and $\displaystyle (x_1,y_1)$ is a point the slope passes through (here it is (pi/2, 1)).

can you continue? - Feb 10th 2008, 11:57 PMearboth
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

$\displaystyle y = 0$

EDIT: According to Jhevons reply the equation of the tangent is $\displaystyle y = 1$ - Feb 11th 2008, 12:04 AMJhevon