Question

derivative of y = sin5x​

Answer 1

Solution:

Given to differentiate:

$\tt\longrightarrow y=sin\:5x$

Differentiating both sides with respect to x, we get:

$\tt\longrightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}(sin\:5x)$

We know that:

$\bigstar\:\:\underline{\boxed{\tt\dfrac{d}{dx}(sin\:nx)=n\:cos\:nx}}$

Using this result, we get:

$\tt\longrightarrow \dfrac{dy}{dx}=5\cdot cos\:5x$

$\tt\longrightarrow \dfrac{dy}{dx}=5\: cos\:5x$

Which is our required answer.

Learn More:

1. Derivative of f(x) using first principle.

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

2. Derivative of standard functions.

$\tt 1.\:\: \dfrac{d}{dx}(k)=0$

$\tt 2.\:\: \dfrac{d}{dx}(x^n)=nx^{n-1}$

$\tt 3.\:\: \dfrac{d}{dx}(a^x)=a^x\:ln\:x$

$\tt 4.\:\: \dfrac{d}{dx}(e^x)=e^x$

$\tt 5.\:\: \dfrac{d}{dx}(log_{a}x)=\dfrac{1}{x\:ln\:a}$

$\tt 6.\:\: \dfrac{d}{dx}(ln\:x)=\dfrac{1}{x}$

$\tt 7.\:\: \dfrac{d}{dx}(|x|)=\dfrac{x}{|x|}$

3. Derivative of trigonometric functions.

$\tt 1.\:\: \dfrac{d}{dx}(sin\:x)=cos\:x$

$\tt 2.\:\: \dfrac{d}{dx}(cos\:x)=-sin\:x$

$\tt 3.\:\: \dfrac{d}{dx}(tan\:x)=sec^2x$

$\tt 4.\:\: \dfrac{d}{dx}(cot\:x)=-cosec^2x$

$\tt 5.\:\: \dfrac{d}{dx}(sec\:x)=sec\:x\:tan\:x$

$\tt 6.\:\: \dfrac{d}{dx}(cosec\:x)=-cosec\:x\:cot\:x$

4. Derivatives of inverse trigonometric functions.

$\tt 1.\:\: \dfrac{d}{dx}(sin^{-1}x)=\dfrac{1}{\sqrt{1-x^2}}$

$\tt 2.\:\: \dfrac{d}{dx}(cos^{-1}x)=\dfrac{-1}{\sqrt{1-x^2}}$

$\tt 3.\:\: \dfrac{d}{dx}(tan^{-1}x)=\dfrac{1}{1+x^2}$

$\tt 4.\:\: \dfrac{d}{dx}(cot^{-1}x)=\dfrac{-1}{1+x^2}$

$\tt 5.\:\: \dfrac{d}{dx}(sec^{-1}x)=\dfrac{1}{|x|\sqrt{x^2-1}}$

$\tt 6.\:\: \dfrac{d}{dx}(cosec^{-1}x)=\dfrac{-1}{|x|\sqrt{x^2-1}}$

5. Derivatives of hyperbolic trigonometric functions.

$\tt 1.\:\: \dfrac{d}{dx}(sinh\:x)=cosh\:x$

$\tt 2.\:\: \dfrac{d}{dx}(cosh\:x)=sinh\:x$

$\tt 3.\:\: \dfrac{d}{dx}(tanh\:x)=sech^2x$

$\tt 4.\:\: \dfrac{d}{dx}(coth\:x)=-cosech^2x$

$\tt 5.\:\: \dfrac{d}{dx}(sech\:x)=-sech\:x\:tanh\:x$

$\tt 6.\:\: \dfrac{d}{dx}(cosech\:x)=-cosech\:x\:coth\:x$

6. Derivatives of inverse hyperbolic trigonometric functions.

$\tt 1.\:\: \dfrac{d}{dx}(sinh^{-1}x)=\dfrac{1}{\sqrt{x^2+1}}$

$\tt 2.\:\: \dfrac{d}{dx}(cosh^{-1}x)=\dfrac{1}{\sqrt{x^2-1}}$

$\tt 3.\:\: \dfrac{d}{dx}(tanh^{-1}x)=\dfrac{1}{1-x^2}$

$\tt 4.\:\: \dfrac{d}{dx}(coth^{-1}x)=\dfrac{1}{1-x^2}$

$\tt 5.\:\: \dfrac{d}{dx}(sech^{-1}x)=\dfrac{-1}{x\sqrt{1-x^2}}$

$\tt 6.\:\: \dfrac{d}{dx}(cosech^{-1}x)=\dfrac{-1}{|x|\sqrt{1+x^2}}$

7. Fundamental rules of derivative.

$\tt 1.\:\: \dfrac{d}{dx}(cf)=c\dfrac{df}{dx}$

$\tt 2.\:\: \dfrac{d}{dx}(f\pm g)=\dfrac{df}{dx}\pm\dfrac{dg}{dx}\:\:\:\:\:\:[Sum\: and\:difference\:\:rules]$

$\tt 3.\:\: \dfrac{d}{dx}(fg)=g\dfrac{df}{dx}+f\dfrac{dg}{dx}\:\:\:\:\:\:[Product\:\:Rule]$

$\tt 4.\:\: \dfrac{d}{dx}\bigg(\dfrac{f}{g}\bigg)=\dfrac{f'g-g'f}{g^2}\:\:\:\:\:\:[Quotient\:\:Rule]$

$\tt 5.\:\: \dfrac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)\:\:\:\:\:\:[Chain\:\:Rule]$

Answer 2

Answer:

Solution:

Given to differentiate:

\tt\longrightarrow y=sin\:5x⟶y=sin5x

Differentiating both sides with respect to x, we get:

\tt\longrightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}(sin\:5x)⟶

dx

dy

=

dx

d

(sin5x)

We know that:

\bigstar\:\:\underline{\boxed{\tt\dfrac{d}{dx}(sin\:nx)=n\:cos\:nx}}★

dx

d

(sinnx)=ncosnx

Using this result, we get:

\tt\longrightarrow \dfrac{dy}{dx}=5\cdot cos\:5x⟶

dx

dy

=5⋅cos5x

\tt\longrightarrow \dfrac{dy}{dx}=5\: cos\:5x⟶

dx

dy

=5cos5x

Which is our required answer.

Learn More:

1. Derivative of f(x) using first principle.

\displaystyle\tt\longrightarrow f'(x)=\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}⟶f

′

(x)=

h→0

lim

h

f(x+h)−f(x)

2. Derivative of standard functions.

\tt 1.\:\: \dfrac{d}{dx}(k)=01.

dx

d

(k)=0

\tt 2.\:\: \dfrac{d}{dx}(x^n)=nx^{n-1}2.

dx

d

(x

n

)=nx

n−1

\tt 3.\:\: \dfrac{d}{dx}(a^x)=a^x\:ln\:x3.

dx

d

(a

x

)=a

x

lnx

\tt 4.\:\: \dfrac{d}{dx}(e^x)=e^x4.

dx

d

(e

x

)=e

x

\tt 5.\:\: \dfrac{d}{dx}(log_{a}x)=\dfrac{1}{x\:ln\:a}5.

dx

d

(log

a

x)=

xlna

1

\tt 6.\:\: \dfrac{d}{dx}(ln\:x)=\dfrac{1}{x}6.

dx

d

(lnx)=

x

1

\tt 7.\:\: \dfrac{d}{dx}(|x|)=\dfrac{x}{|x|}7.

dx

d

(∣x∣)=

∣x∣

x

3. Derivative of trigonometric functions.

\tt 1.\:\: \dfrac{d}{dx}(sin\:x)=cos\:x1.

dx

d

(sinx)=cosx

\tt 2.\:\: \dfrac{d}{dx}(cos\:x)=-sin\:x2.

dx

d

(cosx)=−sinx

\tt 3.\:\: \dfrac{d}{dx}(tan\:x)=sec^2x3.

dx

d

(tanx)=sec

2

x

\tt 4.\:\: \dfrac{d}{dx}(cot\:x)=-cosec^2x4.

dx

d

(cotx)=−cosec

2

x

\tt 5.\:\: \dfrac{d}{dx}(sec\:x)=sec\:x\:tan\:x5.

dx

d

(secx)=secxtanx

\tt 6.\:\: \dfrac{d}{dx}(cosec\:x)=-cosec\:x\:cot\:x6.

dx

d

(cosecx)=−cosecxcotx

4. Derivatives of inverse trigonometric functions.

\tt 1.\:\: \dfrac{d}{dx}(sin^{-1}x)=\dfrac{1}{\sqrt{1-x^2}}1.

dx

d

(sin

−1

x)=

1−x

2

1

\tt 2.\:\: \dfrac{d}{dx}(cos^{-1}x)=\dfrac{-1}{\sqrt{1-x^2}}2.

dx

d

(cos

−1

x)=

1−x

2

−1

\tt 3.\:\: \dfrac{d}{dx}(tan^{-1}x)=\dfrac{1}{1+x^2}3.

dx

d

(tan

−1

x)=

1+x

2

1

\tt 4.\:\: \dfrac{d}{dx}(cot^{-1}x)=\dfrac{-1}{1+x^2}4.

dx

d

(cot

−1

x)=

1+x

2

−1

\tt 5.\:\: \dfrac{d}{dx}(sec^{-1}x)=\dfrac{1}{|x|\sqrt{x^2-1}}5.

dx

d

(sec

−1

x)=

∣x∣

x

2

−1

1

\tt 6.\:\: \dfrac{d}{dx}(cosec^{-1}x)=\dfrac{-1}{|x|\sqrt{x^2-1}}6.

dx

d

(cosec

−1

x)=

∣x∣

x

2

−1

−1

5. Derivatives of hyperbolic trigonometric functions.

\tt 1.\:\: \dfrac{d}{dx}(sinh\:x)=cosh\:x1.

dx

d

(sinhx)=coshx

\tt 2.\:\: \dfrac{d}{dx}(cosh\:x)=sinh\:x2.

dx

d

(coshx)=sinhx

\tt 3.\:\: \dfrac{d}{dx}(tanh\:x)=sech^2x3.

dx

d

(tanhx)=sech

2

x

\tt 4.\:\: \dfrac{d}{dx}(coth\:x)=-cosech^2x4.

dx

d

(cothx)=−cosech

2

x

\tt 5.\:\: \dfrac{d}{dx}(sech\:x)=-sech\:x\:tanh\:x5.

dx

d

(sechx)=−sechxtanhx

\tt 6.\:\: \dfrac{d}{dx}(cosech\:x)=-cosech\:x\:coth\:x6.

dx

d

(cosechx)=−cosechxcothx

6. Derivatives of inverse hyperbolic trigonometric functions.

\tt 1.\:\: \dfrac{d}{dx}(sinh^{-1}x)=\dfrac{1}{\sqrt{x^2+1}}1.

dx

d

(sinh

−1

x)=

x

2

+1

1

\tt 2.\:\: \dfrac{d}{dx}(cosh^{-1}x)=\dfrac{1}{\sqrt{x^2-1}}2.

dx

d

(cosh

−1

x)=

x

2

−1

1

\tt 3.\:\: \dfrac{d}{dx}(tanh^{-1}x)=\dfrac{1}{1-x^2}3.

dx

d

(tanh

−1

x)=

1−x

2

1

\tt 4.\:\: \dfrac{d}{dx}(coth^{-1}x)=\dfrac{1}{1-x^2}4.

dx

d

(coth

−1

x)=

1−x

2

1

\tt 5.\:\: \dfrac{d}{dx}(sech^{-1}x)=\dfrac{-1}{x\sqrt{1-x^2}}5.

dx

d

(sech

−1

x)=

x

1−x

2

−1

\tt 6.\:\: \dfrac{d}{dx}(cosech^{-1}x)=\dfrac{-1}{|x|\sqrt{1+x^2}}6.

dx

d

(cosech

−1

x)=

∣x∣

1+x

2

−1

7. Fundamental rules of derivative.

\tt 1.\:\: \dfrac{d}{dx}(cf)=c\dfrac{df}{dx}1.

dx

d

(cf)=c

dx

df

\tt 2.\:\: \dfrac{d}{dx}(f\pm g)=\dfrac{df}{dx}\pm\dfrac{dg}{dx}\:\:\:\:\:\:[Sum\: and\:difference\:\:rules]2.

dx

d

(f±g)=

dx

df

±

dx

dg

[Sumanddifferencerules]

\tt 3.\:\: \dfrac{d}{dx}(fg)=g\dfrac{df}{dx}+f\dfrac{dg}{dx}\:\:\:\:\:\:[Product\:\:Rule]3.

dx

d

(fg)=g

dx

df

+f

dx

dg

[ProductRule]

\tt 4.\:\: \dfrac{d}{dx}\bigg(\dfrac{f}{g}\bigg)=\dfrac{f'g-g'f}{g^2}\:\:\:\:\:\:[Quotient\:\:Rule]4.

dx

d

(

g

f

)=

g

2

f

′

g−g

′

f

[QuotientRule]

\tt 5.\:\: \dfrac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)\:\:\:\:\:\:[Chain\:\:Rule]5.

dx

d

f(g(x))=f

′

(g(x))⋅g

′

(x)[ChainRule]

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