What is the derivative of $y = sin (tan 2 x)$ $y= sin(tan 2x)$ ?

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What is the derivative of $y = sin (tan 2 x)$ $y= sin(tan 2x)$ ?

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Answer 1 · 2016-01-05T16:40:21

$\frac{d y}{d x} = 2 cos (tan 2 x) {sec}^{2} (2 x)$ $dy/dx = 2cos(tan2x)sec^2(2x)$

Explanation:

We need to apply the chain rule twice.

Recall that the chain rule states, if we have some function $f (g (x))$ $f(g(x))$ , the derivative of $f$ $f$ with respect to $x$ $x$ is equal to the derivative of $f$ $f$ with respect to $g$ $g$ , multiplied by the derivative of $g$ $g$ with respect to $x$ $x$ .

So in this case, the derivative $\frac{d y}{d x}$ $dy/dx$ will equal the derivative of $sin (tan 2 x)$ $sin(tan 2x)$ with respect to $tan 2 x$ $tan 2x$ (basically, treat $tan 2 x$ $tan 2x$ as a whole variable) times the derivative of $tan 2 x$ $tan 2x$ with respect to $x$ $x$ .

Derivative of $sin$ $sin$ is just $cos$ $cos$ :

$\frac{d y}{d x} = cos (tan 2 x) \cdot \frac{d}{d x} [tan 2 x]$ $dy/dx = cos(tan 2x) * d/dx[tan 2x]$

Derivative of $tan$ $tan$ is ${sec}^{2}$ $sec^2$ . However, we need to apply the chain rule again, meaning this time we will just pull the derivative of $2 x$ $2x$ out. (which is just $2$ $2$ )

$\frac{d y}{d x} = cos (tan 2 x) \cdot {sec}^{2} (2 x) \cdot 2$ $dy/dx = cos(tan 2x) * sec^2(2x) * 2$

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What is the derivative of $y = sin (tan 2 x)$ $y= sin(tan 2x)$ ?

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What is the derivative of y= sin(tan 2x)y=sin(tan2x)y= sin(tan 2x)?

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What is the derivative of $y = sin (tan 2 x)$ $y= sin(tan 2x)$ ?