Can you differentiate this please? I keep getting the wrong answers.

Determine the question of the tangent line to y = x^sinx AT x = pi/2

This seemed alot easier when looking at it

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- Apr 5th 2009, 04:26 PM #1

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- Apr 5th 2009, 04:34 PM #2

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$\displaystyle \ln(y) = \ln(x^{\sin(x)}) $

$\displaystyle \ln(y) = \sin(x) \ln(x) $

$\displaystyle \frac{d}{dx} \ln(y) = \frac{d}{dx} \sin(x) \ln(x) $

$\displaystyle \frac{dy}{dx} \times \frac{d}{dy} \ln(y) = \frac{d}{dx} \sin(x) \ln(x)$

$\displaystyle \frac{dy}{dx} \times \frac{1}{y} = \frac{d}{dx} \sin(x) \ln(x)$

Now use the product rule on the RHS, substitude the original equation in for 'y', and rearrange to get $\displaystyle \frac{dy}{dx} = $.

- Apr 5th 2009, 04:34 PM #3
$\displaystyle y = x^{\sin x} \Rightarrow \ln y = \sin x \ln x$.

The derivative of the left hand side is $\displaystyle \frac{1}{y} \cdot \frac{dy}{dx}$. Use the product rule to differentiate the right hand side.

Substitute $\displaystyle x = \frac{\pi}{2}$ and $\displaystyle y = \frac{\pi}{2}$ (why?) and solve for the value of $\displaystyle m = \frac{dy}{dx}$.

You now have the gradient of the tangent and you know a point on the tangent. Getting the equation of the tangent should be routine.

- Apr 5th 2009, 04:37 PM #4
logarithmic derivative ...

$\displaystyle y = x^{\sin{x}}$

$\displaystyle \ln{y} = \ln\left(x^{\sin{x}}\right)$

$\displaystyle \ln{y} = \sin{x} \cdot \ln{x}$

take the derivative of both sides ...

$\displaystyle \frac{y'}{y} = \sin{x} \cdot \frac{1}{x} + \cos{x} \cdot \ln{x}$

$\displaystyle y' = y\left(\sin{x} \cdot \frac{1}{x} + \cos{x} \cdot \ln{x}\right)$

$\displaystyle y' = x^{\sin{x}}\left(\sin{x} \cdot \frac{1}{x} + \cos{x} \cdot \ln{x}\right)

$

sub in $\displaystyle \frac{\pi}{2}$ for x and evaluate.

- Apr 5th 2009, 05:04 PM #5

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- Apr 5th 2009, 05:38 PM #6

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