Question

Find from first principle the derivative of sinx^2 ? step by step explanation​

Answer 1

Answer
First principle of differentiation :
dx
dy
​
=lim
δx→0
​

δx
f(x+δx)−f(x)
​


Here f(x)=sinx

⇒f(x+δx)=sin(x+δx)

⇒f(x+δx)−f(x)=sin(x+δx)−sinx

We know that sinC−sinD=2cos(
2
C+D
​
)sin(
2
C−D
​
)

⇒f(x+δx)−f(x)=2cos(
2
x+δx+x
​
)sin(
2
δx
​
)

⇒
dx
dy
​
=lim
δx→0
​

δx
2cos(x+
2
δx
​
)sin(
2
δx
​
)
​


⇒
dx
dy
​
=lim
δx→0
​
cos(x+
2
δx
​
)
2
δx
​

sin(
2
δx
​
)
​


⇒
dx
d(sinx)
​
=cosx as lim
x→0
​

x
sinx
​
=1
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Answer 2

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:f(x) = sin {x}^{2}$

So,

$\rm :\longmapsto\:f(x + h) = sin {(x + h)}^{2}$

Using definition of First Principle, we have

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

$\rm \:  =  \:  \: \displaystyle\lim_{h \to 0}\tt \dfrac{sin {(x + h)}^{2} - sin {x}^{2} }{h}$

We know,

$\boxed{ \bf{ \: sinx - siny = 2cos\bigg(\dfrac{x + y}{2} \bigg) sin\bigg(\dfrac{x - y}{2} \bigg)}}$

So, using this identity we have

$\rm \: = \:\displaystyle\lim_{h \to 0}\tt \dfrac{2cos\bigg(\dfrac{ {(x + h)}^{2} + {x}^{2} }{2} \bigg)sin\bigg(\dfrac{ {(x + h)}^{2} - {x}^{2} }{2} \bigg)}{h}$

$\rm \:=2\displaystyle\lim_{h \to 0}\tt cos\bigg(\dfrac{ {(x + h)}^{2} + {x}^{2} }{2} \bigg) \times \displaystyle\lim_{h \to 0}\tt \dfrac{sin\bigg(\dfrac{ {(x + h)}^{2} - {x}^{2} }{2} \bigg)}{h}$

$\rm \: =2cos\bigg(\dfrac{ {x}^{2} + {x}^{2} }{2} \bigg) \times \displaystyle\lim_{h \to 0}\tt \dfrac{sin\bigg(\dfrac{ {x}^{2} + {h}^{2} + 2hx - {x}^{2}}{2} \bigg)}{h}$

$\rm \: =2cos\bigg(\dfrac{2{x}^{2}}{2} \bigg) \times \displaystyle\lim_{h \to 0}\tt \dfrac{sin\bigg(\dfrac{{h}^{2} + 2hx }{2} \bigg)}{h}$

$\rm \:=2cos {x}^{2} \times \displaystyle\lim_{h \to 0}\tt \dfrac{sin\bigg(\dfrac{h(h + 2x)}{2} \bigg)}{h}$

$\rm \:=2cos {x}^{2} \times \displaystyle\lim_{h \to 0}\tt \dfrac{sin\bigg(\dfrac{h(h + 2x)}{2} \bigg)}{h\bigg(\dfrac{h + 2x}{2} \bigg)} \times \bigg(\dfrac{h + 2x}{2} \bigg)$

can be rewritten as

$\rm \:=2cos {x}^{2} \times \displaystyle\lim_{h \to 0}\tt \dfrac{sin\bigg(\dfrac{h(h + 2x)}{2} \bigg)}{h\bigg(\dfrac{h + 2x}{2} \bigg)} \times \displaystyle\lim_{h \to 0}\tt \bigg(\dfrac{h + 2x}{2} \bigg)$

We know,

$\boxed{ \bf{ \: \displaystyle\lim_{h \to 0}\tt \frac{sinh}{h} = 1}}$

So, using this identity, we get

$\rm \:  =2cos {x}^{2} \times \bigg(\dfrac{2x}{2} \bigg)$

$\rm \:  =  2x \: cos {x}^{2}$

Hence,

We concluded that,

$\boxed{ \bf{ \: \dfrac{d}{dx} \: sin {x}^{2} \: = \: 2x \: cos {x}^{2}}}$

Additional Information :-

$\boxed{ \bf{ \: \displaystyle\lim_{h \to 0}\tt \frac{tanh}{h} = 1}}$

$\boxed{ \bf{ \: \displaystyle\lim_{h \to 0}\tt \frac{tan {}^{ - 1} h}{h} = 1}}$

$\boxed{ \bf{ \: \displaystyle\lim_{h \to 0}\tt \frac{sin {}^{ - 1} h}{h} = 1}}$

$\boxed{ \bf{ \: \displaystyle\lim_{h \to 0}\tt \frac{log(1 + h)}{h} = 1}}$

$\boxed{ \bf{ \: \displaystyle\lim_{h \to 0}\tt \frac{ {e}^{h} - 1}{h} = 1}}$

$\boxed{ \bf{ \: \displaystyle\lim_{h \to 0}\tt \frac{ {a}^{h} - 1}{h} = loga}}$