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Aditya Kumar 6 years, 3 months ago

Do [tan-1 1 + tan-1 3] + tan-1 2
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Rishabh Rathi 6 years, 3 months ago

It is differentiable in its domain.
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Aditya Kumar 6 years, 3 months ago

Det *

Aditya Kumar 6 years, 3 months ago

Is the value of set A = 9 ?
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S.....Sharma☺??? .. 6 years, 3 months ago

X^3+2x^2-1 is an definate function then at x=1 =(1)^3+2(1)^2-1=1+2-1=2
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Priya Dharshini 6 years, 3 months ago

Understand concept first. Complete ncert exercise each and every sum even solved. Then do 10 yrs sample papers then surely u will top in maths paper..

Abhinav Singh 6 years, 3 months ago

Study

Divyansh Solanki 6 years, 3 months ago

X square DY by DX equal to x square - 2 Y + xy

Aman Kumar 6 years, 3 months ago

Just by making ur concept clear and of lots of practice
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Anubhuti Khatri 6 years, 3 months ago

It was in the book I don't know if it is in syllabus

Priya Dharshini 6 years, 3 months ago

Ohh i am sorry,?i dont know with degree... is this our syllabus question?

Anubhuti Khatri 6 years, 3 months ago

But their a degree also

Priya Dharshini 6 years, 3 months ago

U can do this by first principle derivative....

Vishal Mourya 6 years, 3 months ago

y=cosx dy/dx=dcosx/dx dy/dx= -sinx dx/dx dy/dx= -sinx

Anubhuti Khatri 6 years, 3 months ago

Can you please explain it

Priya Dharshini 6 years, 3 months ago

-sinx...
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Sia ? 6 years, 3 months ago

f(x) = 8x3 and g(x) = {tex}{x^{\frac{1}{3}}}{/tex}
gof = g[f(x)] = g[8x3] {tex}{\left( {8{x^3}} \right)^{\frac{1}{3}}}{/tex} = 2x
and fog = f[g(x)] {tex} = f\left[ {\left( {{x^{\frac{1}{3}}}} \right)} \right] = 8{\left( {{x^{\frac{1}{3}}}} \right)^3}{/tex} = 8x

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S.....Sharma☺??? .. 6 years, 3 months ago

Cos x is an continous function We can prove it by graph because it is continous in graph
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Asvika Mahesh Kumar 6 years, 3 months ago

Because log 2 is constant

Asvika Mahesh Kumar 6 years, 3 months ago

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