ABSTRACT
Using logarithmic differentiation and following the procedure:
(i) Since y= 3e 2θ sec2θ√ (θ −2)
then ln y= ln { 3e2θ sec2θ√
(θ −2) }
= ln ⎧⎨ ⎩3e
⎫⎬ ⎭
(ii) ln y= ln3e2θ + ln sec2θ − ln(θ −2) 1 2
i.e. ln y= ln 3+ ln e2θ + ln sec2θ − 12 ln(θ −2)
i.e. ln y= ln 3+2θ + ln sec2θ − 12 ln(θ −2)
(iii) Differentiating with respect to θ gives: 1 y dy dθ
=0+2+ 2sec2θ tan2θ sec2θ
(θ −2) from equations (1) and (2)
(iv) Rearranging gives: dy dθ
= y { 2+ 2 tan2θ − 1
2(θ − 2) }
(v) dy dθ
= 3e 2θ sec2θ√ (θ − 2)
2+2tan2θ − 1
2(θ − 2) }
Problem 4. Differentiate y= x 3 ln2x
ex sin x with respect
to x
Using logarithmic differentiation and following the procedure gives:
(i) ln y= ln { x3 ln2x ex sin x
}
(ii) ln y= ln x3+ ln(ln2x)− ln(ex)− ln(sin x) i.e. ln y=3ln x + ln(ln2x)− x − ln(sin x)
(iii) 1 y dy dx
= 3 x
ln2x −1− cosx
sin x
(iv) dy dx
= y {
3 x
+ 1 x ln2x
−1− cot x }
(v) dy dx
= x 3 ln2x ex sinx
{3 x
+ 1 x ln2x
−1−cotx }
Now try the following Practice Exercise
Practice Exercise 243 Further problems on differentiating logarithmic functions (answers on page 1137)
In Problems 1 to 6, use logarithmic differentiation to differentiate the given functions with respect to the variable.
