Find the derivative of the given function.f(t)=csc−1(2t+7)f(t)=\csc^{-1}(2t+7)f(t)=csc−1(2t+7)

− 2 ∣ 2 t + 7 ∣ ( 2 t + 7 ) 2 − 1 -\frac{2}{\left|2t+7\right|\sqrt{\left(2t+7\right)^2-1}} − ∣ 2 t + 7 ∣ ( 2 t + 7 ) 2 − 1 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> 2

Find the derivative of the given function. f ( t ) = csc − 1 ( 2 t + 7 ) f(t)=\csc^{-1}(2t+7) f ( t ) = csc − 1 ( 2 t + 7 )

we wanna find the derivative of this function f of t is equal to the inverse cosecant of 2 t+7. So I can see that I have a function inside of my inverse cosecant function, so I know I'm going to have to use the chain rule. So to find our derivative here, f prime of t, starting with that outermost function, the derivative of the inverse cosecant is going to be negative one over the absolute value of whatever we're taking the inverse cosecant of. So that here is 2 t plus 7, then times the square root of that 2 t+7 squared minus 1. Now we're not done yet because using the chain rule, we have to multiply by the derivative of this 2 t+7, which in this case is just 2. And I can write this a bit more compactly if I just write that 2 up top here, and this is my derivative f prime of t. Let us know if you have any questions, and let's get a bit more practice.

Find the derivative of the given function. f ( t ) = csc ⁡ − 1 ⁡ ( 2 t + 7 ) f(t)=\csc^{-1}⁡(2t+7) f ( t ) = csc − 1 ⁡ ( 2 t + 7 )

− 2 ∣ 2 t + 7 ∣ ( 2 t + 7 ) 2 − 1 -\frac{2}{\left|2t+7\right|\sqrt{\left(2t+7\right)^2-1}} − ∣ 2 t + 7 ∣ ( 2 t + 7 ) 2 − 1 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> 2

− 2 ∣ t ∣ t 2 − 1 -\frac{2}{\left|t\right|\sqrt{t^2-1}} − ∣ t ∣ t 2 − 1 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> 2

− 2 ∣ t ∣ ( 2 t + 7 ) 2 − 1 -\frac{2}{\left|t\right|\sqrt{\left(2t+7\right)^2-1}} − ∣ t ∣ ( 2 t + 7 ) 2 − 1 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> 2

− 2 t ∣ 2 t + 7 ∣ ( 2 t + 7 ) 2 − 1 -\frac{2t}{\left|2t+7\right|\sqrt{\left(2t+7\right)^2-1}} − ∣ 2 t + 7 ∣ ( 2 t + 7 ) 2 − 1 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> 2 t

6. Derivatives of Inverse, Exponential, & Logarithmic Functions

Derivatives of Inverse Trigonometric Functions

Calculus

6. Derivatives of Inverse, Exponential, & Logarithmic Functions

Derivatives of Inverse Trigonometric Functions

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Multiple Choice

Find the derivative of the given function.
$f(t)=\csc^{-1}(2t+7)$

$-\frac{2}{\left|t\right|\sqrt{t^2-1}}$

$-\frac{2}{\left|t\right|\sqrt{\left(2t+7\right)^2-1}}$

$-\frac{2t}{\left|2t+7\right|\sqrt{\left(2t+7\right)^2-1}}$

$-\frac{2}{\left|2t+7\right|\sqrt{\left(2t+7\right)^2-1}}$

Verified step by step guidance

Identify the function for which you need to find the derivative: $ f(t) = \csc^{-1}(2t+7) $.

Recall the derivative formula for the inverse cosecant function: $ \frac{d}{dx}[\csc^{-1}(x)] = -\frac{1}{|x|\sqrt{x^2-1}} $.

Apply the chain rule to differentiate $ f(t) = \csc^{-1}(2t+7) $. The chain rule states that $ \frac{d}{dt}[\csc^{-1}(u)] = \frac{d}{du}[\csc^{-1}(u)] \cdot \frac{du}{dt} $, where $ u = 2t+7 $.

Differentiate $ u = 2t+7 $ with respect to $ t $, which gives $ \frac{du}{dt} = 2 $.

Substitute $ u = 2t+7 $ and $ \frac{du}{dt} = 2 $ into the chain rule expression: $ \frac{d}{dt}[\csc^{-1}(2t+7)] = -\frac{2}{|2t+7|\sqrt{(2t+7)^2-1}} $.

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