How do you differentiate $y = csc θ (θ + cot θ)$ $y=csctheta(theta+cottheta)$ ?

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How do you differentiate $y = csc θ (θ + cot θ)$ $y=csctheta(theta+cottheta)$ ?

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Answer 1 · 2017-11-22T14:00:50

$\frac{d y}{d θ} = - csc θ (θ cot θ + {cot}^{2} θ - 1 + {csc}^{2} θ)$ $(dy)/(d theta)=-csctheta(thetacottheta+cot^2theta-1+csc^2theta)$

Explanation:

we will need to use the product rule

$\frac{d}{d x} (u v) = v \frac{d u}{d x} + u \frac{d v}{d x}$ $d/(dx)(color(red)(u)color(blue)(v))=color(blue)(v)color(red)((du)/(dx))+color(red)(u)color(blue)((dv)/(dx))$

$y = csc θ (θ + cot θ)$ $y=csctheta(theta+cottheta)$

$\frac{d y}{d θ} = (θ + cot θ) \frac{d}{d θ} (csc θ) + csc θ \frac{d y}{d θ} (θ + cot θ)$ $(dy)/(d theta)=color(blue)((theta+cottheta))color(red)(d/(d theta)(csctheta))+color(red)(csctheta)color(blue)((dy)/(d theta)(theta+cottheta)$

$\frac{d y}{d θ} = (θ + cot θ) (- csc θ cot θ) + csc θ (1 - {csc}^{2} θ)$ $(dy)/(d theta)=(theta+cottheta)(-cscthetacottheta)+csctheta(1-csc^2theta)$

tiding up.

$\frac{d y}{d θ} = - csc θ (θ cot θ + {cot}^{2} θ - 1 + {csc}^{2} θ)$ $(dy)/(d theta)=-csctheta(thetacottheta+cot^2theta-1+csc^2theta)$

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How do you differentiate $y = csc θ (θ + cot θ)$ $y=csctheta(theta+cottheta)$ ?

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Science

Math

Humanities

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How do you differentiate y=cscθ(θ+cotθ)y=csctheta(theta+cottheta)?

1 Answer

Explanation:

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How do you differentiate $y = csc θ (θ + cot θ)$ $y=csctheta(theta+cottheta)$ ?