tangent line

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February 10th 2008, 11:24 PM
Bust2000

tangent line

find the equation of the tangent line to y=(sin x)^4x at the point (pi/2, 1)
February 10th 2008, 11:55 PM
Jhevon

Quote:

Originally Posted by Bust2000

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

a line is of the form $y = mx + b$

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

can you continue?
February 10th 2008, 11:57 PM
earboth

Quote:

Originally Posted by Bust2000

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

Do you really mean:

$f(x) = \left(\sin(x)\right)^{4x}$ ...... ??

If so re-write f to:

$f(x) = e^{4x \cdot \ln(\sin(x))}$ ...... and use the chain rule to calculate the derivative:

$f'(x) = \frac{e^{4x \cdot \ln(\sin(x))}}{\sin(x)} \cdot \left(4 \sin(x) \cdot \ln(\sin(x)) + 4x \cdot \cos(x) \right)$

Now plug in $x = \frac{\pi}{2}$ to calculate the slope of the tangent.

Since $\ln\left(\sin\left(\frac{\pi}2 \right) \right) = 0$ and $\cos \left(\frac{\pi}2 \right) = 0$ too you'll get:

$f'\left(\frac{\pi}2 \right) = 0$ and therefore the equation of the tangent is

$y = 0$

EDIT: According to Jhevons reply the equation of the tangent is $y = 1$
February 11th 2008, 12:04 AM
Jhevon

Quote:

Originally Posted by earboth

Do you really mean:

$f(x) = \left(\sin(x)\right)^{4x}$ ...... ??

If so re-write f to:

$f(x) = e^{4x \cdot \ln(\sin(x))}$ ...... and use the chain rule to calculate the derivative:

$f'(x) = \frac{e^{4x \cdot \ln(\sin(x))}}{\sin(x)} \cdot \left(4 \sin(x) \cdot \ln(\sin(x)) + 4x \cdot \cos(x) \right)$

Now plug in $x = \frac{\pi}{2}$ to calculate the slope of the tangent.

Since $\ln\left(\sin\left(\frac{\pi}2 \right) \right) = 0$ and $\cos \left(\frac{\pi}2 \right) = 0$ too you'll get:

$f'\left(\frac{\pi}2 \right) = 0$ and therefore the equation of the tangent is

$y = 0$

i believe it would be y = 1 if the slope is zero