Math, asked by pradeepkumar7, 1 year ago

d by dx of f(x)
differentiation chapter

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Answers

Answered by saurabhsemalti
0
f(x)=sin.^2 (x^3+x^2+x+1)^2
df(x)/dx=2(x^3+x^2+x+1)(3x^2+2x+1)[sin2(x^3+x^2+x+1)^2]
here is answer...... also differentiation of sin sq. x =sin2x
Answered by ShresthaTheMetalGuy
0

Given:

f(x)=sin²(x³+x²+x+1)

Now, Let(also),

t=(x³+x²+x+1)²

» f(x)=sin²t

To Find:

 \frac{d}{dx} [f(x)]

Solution:

On applying the chain rule, i.e.,

 \frac{d[f(x)]}{dx}  =  \frac{d[f(x)]}{dt}  \times  \frac{d[t]}{dx}

Thus,

  \frac{d \:[ f(x)]}{dt} =  \frac{d}{dt} [ \sin {}^{2} (t) ]

 \frac{d \: [f(x)]}{dt} = 2 \sin(t) . \cos(t)   =  \sin(2t)

And,

 \frac{d \: t}{dx}  = \frac{d}{dx} [(x {}^{3} + x {}^{2}  + x + 1) {}^{2}  ]

 \frac{d \: t}{dx}  = 6x {}^{5 }  + 10x {}^{4}  + 12x {}^{3}  + 12x {}^{2}  + 6x + 2

Now, On mutliplying both the derivatives:

 \frac{d \:[ f(x)]}{dx} =  \sin(2t) (6x {}^{5 }  + 10x {}^{4}  + 12x {}^{3}  + 12x {}^{2}  + 6x + 2)

 =  \sin(2(x {}^{3} + x {}^{2}  + x + 1) {}^{2}  ) (6x {}^{5 }  + 10x {}^{4}  + 12x {}^{3}  + 12x {}^{2}  + 6x + 2)

 =  2\sin(2(x {}^{6} + 2x {}^{5}  + 3x {}^{4}   + 4x {}^{3} + 3x {}^{2}  + 2x + 1  ) )(3x {}^{5 }  + 5x {}^{4}  + 6x {}^{3}  + 6x {}^{2}  + 3x + 1)

So, the derivative of f(x) w.r.t x is given by:

» " 2sin[2{(x+1)(x²+1)}²].(x+1).(x²+1).(3x²+2x+1) "

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