Question

Differentiate acos³x step by step.

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Answer 1

EXPLANATION.

Differentiation of acos³x.

As we know that,

In this type of question, first we differentiate power then constant term,

⇒ dy/dx = d(acos³x)/dx.

⇒ dy/dx = 3acos²x(-sinx).

⇒ dy/dx = -3acos²x(sin x).

                                                                                         

MORE INFORMATION.

Differentiation by trigonometrical substitutions.

(1) = √a² - x²

Substitution x = a sin∅  or  a cos∅.

(2) = √x² + a²

Substitution x = a tan∅  or  a cot∅.

(3) = √x² - a²

Substitution x = a sec∅  or  a cosec∅.

(4) = √a - x/a + x

Substitution x = a cos2∅.

(5) = √a² - x²/a² + x²

Substitution x² = a² cos 2∅.

(6) = √ax - x².

Substitution x = a sin²∅.

(7) = √x/a + x

Substitution x = atan²∅.

(8) = √x/a - x.

Substitution x = a sin²∅.

(9) = √(x - a)(x - b).

Substitution x = a sec²∅ - b tan²∅.

(10) = √(x - a)(b - x)

Substitution x = acos²∅ + b sin²∅.

Answer 2

${\large{\bold{\rm{\underline{Required \; Solution}}}}}$

$\; \; \; \; \;{\bullet}$ Differentiate acos³x

$\; \; \; \; \;{\bullet}$ dx/dy = a3cos²x(-sin x)

$\; \; \; \; \;{\bullet}$ dx/dy = -3a cos²x(sin x)

${\large{\bold{\rm{\underline{Additional \; information}}}}}$

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• d(e^x)/dx = e^x

• d(x^n)/dx = n x^(n-1)

• d(ln x)/dx = 1/x

• d(sin x)/dx = cos x

• d(cos x)/dx = - sin x

• d(tan x)/dx = sec² x

• d(sec x)/dx = tan x * sec x

• d(cot x)/dx = - cosec²x

• d(cosec x)/dx = - cosec x * cot x

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Trigonometry Table -

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}$

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Some important trigonometry identity -

• cosec²θ - cot²θ = 1

• cosec²θ = 1 + cot²θ

• 1 + cot²θ = cosec²θ

• sin²θ + cos²θ = 1

• sin²θ = 1 - cos²θ

• cos²θ = 1 - sin²θ

• sec²θ = 1 + tan²θ

• sec²θ - tan²θ = 1

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