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2 Answers
- LearnerLv 71 decade agoFavorite Answer
Let 'y = (7t)^sin(t)'
Taking logarithms on both sides,
ln(y) = sin(t)*ln(7t)
Differentiating, with respect to 't',
(1/y)*(dy/dt) = cos(t)*ln(7t) + sin(t)*(1/7t)*(7),
which simplifies to:
(1/y)*(dy/dt) = cos(t)*ln(7t) + sin(t)*(1/t)
==> dy/dt = (y)[cos(t)*ln(7t) + sin(t)/(t)]
==> dy/dt = (7t)^(sin t)[cos(t)*ln(7t) + sin(t)/(t)]
- Anonymous1 decade ago
y = 7t ^ sin t
ln y = (sin t)(ln 7t)
Differentiate
1/y dy/dt = (cos t) (ln 7t) + (sin t) (1/t)
dy/dt = [ cos t ln 7t + (1/t) sin t ] 7t ^ sin t
