Find the derivative of the given function.y=tan(e−x3)y=\tan\left(e^{-x^3}\right)

−3x2⋅e−x3⋅sec⁡2(e−$$x3)-3x^2$$\cdot $$e^{-x^3}$$\cdot\$$sec^2$$\$$left(e^{-x^3}$$\right)−3x2⋅e−x3⋅sec2(e−x3)

Find the derivative of the given function. y = tan ( e − x 3 ) y=\tan\left(e^{-x^3}\right)

In this problem, we want to find the derivative of the function y equals tangent of e to the power of negative x cubed. So let's go ahead and get into things here. We want to find our derivative here, y prime, or you could also indicate that as dydx, doesn't matter, means the same thing. Now starting with our outermost function here because we're obviously going to have to use the chain rule because I have a function that has another function inside that has another function inside. So going ahead and working with that outermost function, the derivative of tangent is the secant squared. So this is the secant squared of e to the power of negative x cubed. Then multiplying by the derivative, working my way inside the derivative of e to the power of negative x cubed, This is going to be e to the power of negative x cubed times the derivative of that negative x cubed, which is, of course, negative three x squared using the power rule. Now we're gonna rearrange this a little bit just to give this a little bit more standard. Look. This is negative 3x squared e times e to the power of negative x cubed times the secant squared of e to the power of negative x cube. So this is our final derivative here, y prime of x, having applied the chain rule along with all of our other rules for derivatives. Feel free to let us know if you have any questions, and let's keep practicing.

Find the derivative of the given function.y=tan⁡(e−x3)y=\tan\$$left(e^{-x^3}$$\right)

−3x2⋅e−x3⋅sec⁡2(e−$$x3)-3x^2$$\cdot $$e^{-x^3}$$\cdot\$$sec^2$$\$$left(e^{-x^3}$$\right)−3x2⋅e−x3⋅sec2(e−x3)

sec⁡2(e−x3)\$$sec^2$$\$$left(e^{-x^3}$$\right)sec2(e−x3)

e−x3⋅sec⁡2(e−$$x3)e^{-x^3}$$\cdot\$$sec^2$$\$$left(e^{-x^3}$$\right)e−x3⋅sec2(e−x3)

−x3⋅e−x3−1⋅sec⁡2(e−$$x3)-x^3$$\cdot $$e^{-x^3-1}$$\cdot\$$sec^2$$\$$left(e^{-x^3}$$\right)−x3⋅e−x3−1⋅sec2(e−x3)

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6. Derivatives of Inverse, Exponential, & Logarithmic Functions

Derivatives of Exponential & Logarithmic Functions

Calculus

6. Derivatives of Inverse, Exponential, & Logarithmic Functions

Derivatives of Exponential & Logarithmic Functions

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Find the derivative of the given function.
$y = \tan (e^{- x^{3}})$

$$\sec$^2$\left$(e^{-x^3}$\right$)$

$e^{-x^3}$\cdot\]\sec$^2$\left$(e^{-x^3}$\right$)$

$-3x^2$\cdot$ e^{-x^3}$\cdot\]\sec$^2$\left$(e^{-x^3}$\right$)$

$-x^3$\cdot$ e^{-x^3-1}$\cdot\]\sec$^2$\left$(e^{-x^3}$\right$)$

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Identify the function y = $\tan$$\left$(e^{-x^3}$\right$). We need to find its derivative with respect to x.

Recognize that this is a composite function, where the outer function is $\tan$(u) and the inner function is u = e^{-x^3}. We will use the chain rule to differentiate.

Differentiate the outer function $\tan$(u) with respect to u, which gives $\sec$^2(u).

Differentiate the inner function u = e^{-x^3} with respect to x. This requires using the chain rule again: differentiate e^{-x^3} to get -3x^2 $\cdot$ e^{-x^3}.

Combine the derivatives using the chain rule: multiply $\sec$^2(e^{-x^3}) by the derivative of the inner function, -3x^2 $\cdot$ e^{-x^3}, to get the derivative of the original function.