Question

diferentite sin 3x with respect to x​

Answer 1

$\sf \dfrac{d}{dx} sin^3 x$

• use chain rule,

$\sf 3 = \dfrac{d}{dx} sin(x) \times sin^2 x$

• again use chain rule,

$\sf = cos x \dfrac{d}{dx} x \ 3 \ sin^2 x$

the derivative of x is 1

$\sf = 1 \times 3 \ cos x \ sin^2 x$

$\sf = 3 \ cos x \ sin^2 x$

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differential equation Identities

$\boxed{\boxed{\begin{minipage}{5cm}\displaystyle\circ\sf\;\dfrac{d}{dx}(sin\;x)=cosx \\\\ \circ \;\dfrac{d}{dx}(cos\;x) = -sinx \\\\ \circ \; \dfrac{d}{dx}(tan\;x) = sec^{2}x \\\\ \circ\; \dfrac{d}{dx}(cot\;x) = -csc^{2}x \\\\ \circ \; \dfrac{d}{dx}(sec\;x) = secx \cdot tanx \\\\ \circ \; \dfrac{d}{dx}(csc\;x) = -cscx \cdot cotx \\\\ \circ\; \dfrac{d}{dx}(sinh\;x)=coshx \\\\ \circ\; \dfrac{d}{dx}(cosh\;x)= sinhx \\\\ \circ\;\dfrac{d}{dx}(tanh\;x)=sech^{2}h \\\\ \circ\;\dfrac{d}{dx}(coth\;x)=-csch^{2}x \\\\ \circ\;\dfrac{d}{dx}(sech\;x) =-sechx \cdot tanhx \\\\ \circ\;\dfrac{d}{dx}(csch\;x) = -cschx \cdot cothx\end{minipage}}}$

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Answer 2

‌

‌Answer :

• Derivative of sin 3x with respect to x is 3cos3x.
• Derivative of sin³x with respect to x is 3sin²x cosx.

‌

Solution :

‌➽ Differentiation of sin3x with respect to x

‌$\large\bf \blue{\underbrace{\red{Method \:1}}}:$

‌Using First principle of Derivative.

‌

‌ $f'(x) = \displaystyle{\lim_{h\to 0}}\frac{f(x+h)-f(x)}{h}$

‌Let f(x) = sin3x. Then

‌$f'(x) = \displaystyle{\lim_{h\to 0}}\frac{sin3(x+h)-sin3(x)}{h} \\\\ \:\:\: = \displaystyle{\lim_{h\to 0}} \frac{sin(3x+3h)-sin3x}{h}$

‌$\red\bigstar\boxed{\sf sin x - sin y = 2cos \frac{x+y}{2}sin \frac{x-y}{2}}$

‌

‌Applying this formula here, we got :

‌$\sf = \displaystyle{\lim_{h\to 0}}\frac{2cos\frac{(3x+3h+3x)}{2} sin \frac{(\cancel{3x}+3h\:\:\cancel{-3x} )}{2}}{h} \\\\ = \sf\displaystyle{\lim_{h\to 0}}\frac{2cos\frac{(6x+3h)}{2} sin \frac{3h }{2}}{h} \\\\ \sf = \displaystyle{\lim_{h\to 0}}\:\frac{2cos(3x + \frac{3h}{2}) sin \frac{3h }{2}}{h}$

‌Multiplying and dividing by 3/2

‌$\sf = \displaystyle{\lim_{h\to 0}}\frac{2cos(3x + \frac{3h}{2}) sin \frac{3h }{2}\times \red{\frac{3}{2}}}{h\times \red{\frac{3}{2}}}$

‌We know that,

‌$\sf \blue{\bigstar} \boxed{\color{red} \displaystyle{\lim_{h\to 0}}\:\sf\frac{sin x}{x}= 1}$

‌$\sf = \displaystyle{\lim_{h\to 0}}\sf\frac{2cos(3x + \frac{3h}{2}) \blue{sin \frac{3h }{2}} \times \frac{3}{2}}{\blue{\frac{3}{2}h}}$

‌

‌$\sf Putting \: \green{ \displaystyle{\lim_{h\to 0}}\:\frac{sin \frac{3h}{2}}{\frac{3h}{2}}= 1}$

‌

‌$\sf \displaystyle{\lim_{h\to 0}}\:\:\sf \cancel{2}cos (3x + \frac{3h}{2}) \times \frac{3}{\cancel{2}} \\\\ \sf \displaystyle{\lim_{h\to 0}} \sf\:\:3cos (3x + \frac{3h}{2})$

‌

‌Putting h = 0

‌$\sf = 3cos (3x + \frac{3\times (0)}{2}) \\ \\ \large\implies \boxed{\sf 3cos (3x )}$

‌

‌━━━━━━━━━━━━━━━━━

‌$\large\bf \blue{\underbrace{\red{Method\: 2}}}:$

‌Chain rule :

‌According to chain rule :

‌Derivative of F(x) is F'(x)

‌F'(x) = f'(g(x)).g'(x)

‌Here, f(x) = sin(3x) and g(x) = 3x

‌ = f'(g(x)).g'(x)

‌= (sin(3x))' .(3x)'

‌= cos 3x . 3

‌= 3cos 3x

‌

Hence,

Derivative of sin 3x with respect to x is 3cos 3x.

‌━━━━━━━━━━━━━━━━━━━━━

‌

‌➽ Differentiation of sin³x with respect to x

‌Using chain rule :

‌Let y = sin³x

‌$\sf \dfrac{d}{dx}sin^3x \times \dfrac{d}{dx}sin x$

‌We know that,

‌$\sf \: Derivative \: of \:\boxed{\sf \color{red} x^n = n.x^{n-1}}$

‌$\sf \: Derivative \: of \:\boxed{\sf \color{red} sin x = cos x}$

‌Applying this here, we got :

‌$\sf 3sin^{3-1} \times cos x \\ \\ \implies\sf 3sin^2 x cos x$

‌

‌Hence,

Derivative of sin³x with respect to x is 3sin²x cosx.

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