What is the derivative of $f (t) = (t {cos}^{2} t, t^{2} - t sec t)$ $f(t) = (tcos^2t , t^2-t sect )$ ?

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What is the derivative of $f (t) = (t {cos}^{2} t, t^{2} - t sec t)$ $f(t) = (tcos^2t , t^2-t sect )$ ?

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Answer 1 · 2017-11-24T07:17:48

$\frac{d y}{d x} = \frac{2 t - sec t - t sec t tan t}{{cos}^{2} t - t sin 2 t}$ $(dy)/(dx)=(2t-sect-tsect tant)/(cos^2t-tsin2t)$

Explanation:

When a functiion is given in parametric form such as $f (t) = (x (t), y (t))$ $f(t)=(x(t),y(t))$ , its dervative is given by $\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}}$ $(dy)/(dx)=((dy)/(dt))/((dx)/(dt))$

Here we have $y (t) = t^{2} - t sec t$ $y(t)=t^2-tsect$ hence $\frac{d y}{d t} = 2 t - sec t - t sec t tan t$ $(dy)/(dt)=2t-sect-tsect tant$

and $x (t) = t {cos}^{2} t$ $x(t)=tcos^2t$ hence $\frac{d x}{d t} = {cos}^{2} t + t \times 2 cos t \times (- sin t)$ $(dx)/(dt)=cos^2t+txx2costxx(-sint)$

= ${cos}^{2} t - t sin 2 t$ $cos^2t-tsin2t$

Hence $\frac{d y}{d x} = \frac{2 t - sec t - t sec t tan t}{{cos}^{2} t - t sin 2 t}$ $(dy)/(dx)=(2t-sect-tsect tant)/(cos^2t-tsin2t)$

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What is the derivative of $f (t) = (t {cos}^{2} t, t^{2} - t sec t)$ $f(t) = (tcos^2t , t^2-t sect )$ ?

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What is the derivative of f(t)=(tcos2t,t2−tsect)f(t) = (tcos^2t , t^2-t sect ) ?

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What is the derivative of $f (t) = (t {cos}^{2} t, t^{2} - t sec t)$ $f(t) = (tcos^2t , t^2-t sect )$ ?