If (sin x)ʸ = yᶜᵒˢˣ, then find dy/dx.
differential calculus topic of logarthimic differentiation
given
(sin x)ʸ = yᶜᵒˢˣ [power contains variables]
Take ln both sides
ln[(sin x)ʸ] = ln[yᶜᵒˢˣ] [use properties of logarithms]
y. ln(sin x) = cos x . ln y
Differentiate w.r.t. x
[use product rule and chain rule]
[dy/dx] ln(sin x) + y cot x = −sin x · ln y + cos x · (1/y) ·[ dy/dx]
[dy/dx ] ln(sin x) + y cot x = −sin x ln y + (cos x / y) [dy/dx]
Collect dy/dx terms
[dy/dx] ln(sin x) − (cos x / y)[ dy/dx ]= −sin x ln y − y cot x
[ dy/dx] [ln(sin x) − cos x / y] = −[sin x ln y + y cot x]
dy/dx = −[sin x ln y + y cot x] / [ln(sin x) − cos x / y]
Multiply numerator and denominator by y:
dy/dx = −y[sin x ln y + y cot x] / [y ln(sin x) − cos x]
see this video for more explanation
No comments:
Post a Comment
please leave your comments