prove that d/dx[x/2×(√a^2-x^2)+a^2/2×sin^-1x/a]=√a^2-x^2
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d/dx[x/2(√(a²-x²))]= 1/2*√(a²-x²) + x/2*(-2x/2√(a²-x²))
and
d/dx[a²/2*sin^-1(x/a)]=
a²/2 * 1/√[1-(x/a)²]*(1/a)
add both you will get the answer
and
d/dx[a²/2*sin^-1(x/a)]=
a²/2 * 1/√[1-(x/a)²]*(1/a)
add both you will get the answer
Harsh1113:
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