Mathematics 1G1. Autumn 2016
Theodore Voronov.
MATH19731. Autumn semester 2016-2017
Exercise sheet 3 (Differentiation-1). Solutions.
Below are the answers for the differentials dy; the answers for the derivatives dy/dx can
be obtained by dividing by dx. It is advisable to calculate the differentials directly, for
example: d(2x ) = d(ex ln 2 ) = (ln 2)2x dx.
1. dy = (2x ln(2) + 3x ln(3))dx,
2. dy =
dx
dx
−
x ln(2) x ln(5)
3. dy = (2x + 1) cos(x) + 2 sin(x) dx
4. dy = ex (sec2 (x) + ex tan(x)) dx
dx
cos(x)
sin(x)
x
−
dx
5. dy =
= −
(ln(x))2
ln(x)
x(ln(x))2
(sin(x) + cos(x))(8x3 ) − (2x4 )(cos(x) − sin(x))
6. dy =
dx
(sin(x) + cos(x))2
ln(x) · (− sin(x)dx) − cos(x) ·
7. dy = d ((2x + 1)3 ) = 3(2x + 1)2 d(2x + 1) = 3(2x + 1)2 · 2dx = 6(2x + 1)2 dx
√
−6xdx
3xdx
d(4 − 3x2 )
= √
= −√
8. dy = d 4 − 3x2 = √
2 4 − 3x2
2 4 − 3x2
4 − 3x2
√
√
√
√
9. dy = d sin(x + x) = cos(x2 + x) d(x2 + x) = cos(x2 + x)
√
1
2
dx
cos(x + x) 2x + √
2 x
2
2
2
dx
2xdx + √
=
2 x
2
10. dy = d(e−x ) = e−x d(−x2 ) = −2xe−x dx
d(x2 + 2x)
(2x + 2)dx
(x + 1)dx
1
√
√
√
=−
=−
=−
11. dy = d √
2
2
2
2
2
2
x + 2x
2(x + 2x) x + 2x
2(x + 2x) x + 2x
(x + 2x) x2 + 2x
12. dy = d ln(cos(2x)) =
d cos 2x
−2 sin 2x dx
=
= −2 tan 2x dx
cos 2x
cos 2x
13. dy = d tan4 (5x) = 4 tan3 (5x) d tan(5x) = 4 tan3 (5x)
1
20 sin3 (5x)dx
5dx
=
cos2 5x
cos5 (5x)
Mathematics 1G1. Autumn 2016
Theodore Voronov.
14. ln(y) = x ln(x2 + 3); by differentiating both sides,
dy
2xdx
2x2
2
2
= dx ln(x + 3) + x 2
= ln(x + 3) + 2
dx
y
x +3
x +3
2x2
2x2
2
2
2
x
Hence, dy = y ln(x + 3) + 2
dx = (x + 3) ln(x + 3) + 2
dx .
x +3
x +3
15. ln(y) = 3x ln(tan(x)); by differentiating both sides,
dy
3xdx
x
= 3dx ln(tan(x)) +
= 3 ln(tan(x)) +
dx
y
tan(x) cos2 x
sin(x) cos(x)
x
3x
dx .
Hence dy = 3(tan(x))
ln(tan(x)) +
sin(x) cos(x)
2