Question

differentiation of sin^3x​

Answer 1

Answer:

Basic formula:

d

d

x

x

n

=

n

x

n

−

1

d

d

x

(

sin

x

)

=

cos

x

Now, let's move to the question:

=

d

d

x

(

sin

3

x

)

=

(

3

sin

2

x

)

×

(

d

d

x

(

sin

x

)

)

=

3

sin

2

x

cos

x

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Answer 2

To find :

Differentiation of sin³x.

Solution :

We know the composite function rule i.e,

$\boxed{\bf{\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}}}$

Here in our case ,

• dy = sin²x
• du = sinx

Using the Composite rule and by substituting the values in it, we get :

$:\implies \bf{\dfrac{dy}{dx} = \dfrac{d(sin^{3}x}{d(sin x)} \times \dfrac{d(sin x)}{dx}}$

$:\implies \bf{\dfrac{dy}{dx} = \dfrac{d(sin^{3}x)}{d(sin x)} \times \dfrac{d(sin x)}{dx}}$

We know that the Differentiation of d(sin x)/d(x) is cos x.

By substituting it in the equation, we get :

$:\implies \bf{\dfrac{dy}{dx} = \dfrac{d(sin^{3}x}{d(sin x)} \times cos x}$

Now by differentiating , d(sin³x)/d(sin x) by derivative by first principle , we get :

$\boxed{\bf{f'(x) = \lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h}}}$

$:\implies \bf{f'(x) = \lim_{h \to 0} \dfrac{sin(x + h)^{3} - sin^{3}x}{h}}$

By using the identity , (a + b)³ = a³ + b³ + 3a²b + 3ab² , we get :

$:\implies \bf{f'(x) = \lim_{h \to 0} \dfrac{sin(x + h)^{3} - sin^{3}x}{h}}$

$:\implies \bf{f'(x) = \lim_{h \to 0} \dfrac{sin(x^{3} + h^{3} + 3x^{2}h + 3xh^{3}) - sin^{3}x}{h}}$

$:\implies \bf{f'(x) = \lim_{h \to 0} \dfrac{sin x^{3} + sin h^{3} + 3x^{2}h + 3sinxh^{3} - sin^{3}x}{h}}$

$:\implies \bf{f'(x) = \lim_{h \to 0} \dfrac{sin h^{3} + 3sinx^{2}h + 3sinxh^{3}}{h}}$

$:\implies \bf{f'(x) = \lim_{h \to 0} sin^{2}h + 3sin^{2}x + 3sinxh^{2}}$

$:\implies \bf{f'(x) = sin 0^{2} + 3sin^{2}x + sin3x(0)^{2}}$

$:\implies \bf{f'(x) = 3sin^{2}x}$

Hence the Differentiation of d(sin³x)/d(sin x) is 3sin²x

Now putting it in the equation ,

$\bf{\dfrac{dy}{dx} = \dfrac{d(sin^{3}x)}{d(sin x)} \times cos x}$

We get ,

$:\implies \bf{\dfrac{dy}{dx} = 3sin^{2}x \times cos x}$

$:\implies \bf{\dfrac{dy}{dx} = 3sin^{2}xcos x}$

$\boxed{\therefore \bf{\dfrac{dy}{dx} = 3sin^{2}xcos x}}$

Hence the Differentiation of sin³x is 3sin²xcosx.