1.i. "ln [ 1 + sin 2x]"
Differentiating with respect to "x" :
"d( ln [ 1 + sin2x])\/dx"
"=[2 cos(2x)] \/ [ 1 + sin 2x]" (Answer)
ii. "x^x"
Let A = "x^x"
Taking natural logarithm on both sides:
"ln A= x ln x"
Differentiating with respect to x:
"d[ ln A]\/dx = d[x ln x] \/ dx"
or, "1 \/ A * d[A]\/dx = x\/x + ln x"
or, "dA\/dx = (1 + ln x) * A"
Putting A = "x^x"
"dA\/dx = [1 + ln x] * X^x Answer"
c. Solve "\\int ( 2x^3 - 4x - 8\/x^4 - x^3 + 4x^2 -4x) dx"
= "2x^4\/4 - 4x^2\/2 - 8x^{-3} \/( -3) - x^4\/4 + 4x^3\/3 - 4x^2\/2 + c"
where C is an integration constant
= "x^4\/2 - 2x^2 + 8(x^-3 \/ 3) - x^4\/4 + 4x^3\/3 - 2x^2 + c"
="[x^4\/2 - x^4\/4] - [ 2x^2 + 2 x^2] + 8x^-3\/3 + 4x^3\/3 + c"
="[x^4\/4] - 4x^2 + 8\/3x^3 + 4x^3\/3 + c [Answer]"
Comments
Leave a comment