Question

a) Find f'(x) using logarithmic differentiation, where f(x) = e−3x√2x−5 (6−5x)4

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(d) Evaluate the integralZ(x3 + 1)1/3x5dx.

Answer 1

a. f(x) = exp(-3$x(2x)$ ^0.5 $- 5(6-5x)4$ )

Taking natural logarithm on both sides:

$ln f(x) = -3x(2x)$ ^0.5 - $20(6-5x)$

or, $ln f(x) = -3x(2x)$ ^0.5 - $120 + 100x$

$=-3(2)$ ^0.5$x$ ^(1.5) $- 120 + 100x$

Differentiating w.r.t $x:$ $-(1/f(x)) * d(f(x))/dx$ $= -3*(2)$ ^0.5 * $x$ ^(1.5+1)$/(1.5+1)$ $+100$

or, $(1/f(x))*d(f(x))/dx=( -3*(2)$ ^0.5 * ($x)$ ^(2.5) )*($2/5)$ +$100$

or, $d(f(x))/dx =$ $[(-6*2$ ^0.5 * $x$ ^2.5)$/5 + 100]$ * exponent($-3x * ($ $2x)$ ^0.5 $-20[6-5x]$ (Answer)

b) $Z= \int$ $(x$ ³$+1)/(3x^5)$ $dx$

$= \int[x^3/(3x^5) + 1/(3x^5)]dx$

$=\int[1/(3x^2) + 1/(3x^5)]dx$

$=(1/3) \int[ 1/x^2+ 1/x^5]dx$

$=(1/3) [x$ ^(-2+1)/(-2+1) + $x$ ^(-5+1)/(-5+1)] + C

$=(1/3)[-(1/x) - 1/(4x^4)] + c$

$=-(1/3)[(1/x) + 1/(4x^4)] +c$ (Answer)