What is the d/dx x^tanx?
y=x^tanx
taking log on both side
⇒logy=log(x^tanx)
or
⇒logy=tanx.logx
⇒differentiate both side
⇒(1/y)y’=tanx.(1/x)+logx.sec^2x
⇒y’={tanx/x+(logx.sec^2x)}*y
⇒y’={tanx/x+(logx.sec^2x)}*x^tanx
⇒y’={tanx/x+(logx.sec^2x)}*x^tanx
What is
This function can be written as
This function can be written as
y=xy
take log on both sides
logy=logxy
logy=ylogx
diffentiate on both sides
1/y(dy/dx)=y/x+logx(dy/dx)
dy/dx(1/y-logx)=y/x
dy/dx=(y/x)/(1/y-logx)
What is the value of the derivative of log (base a) x?
y=logax
What is the value of the derivative of log (base a) x?
y=logax
y=logx/loga
dy/dx=(1/x)/loga
dy/dx=logae/x
What is the differentiation of ?
y=ln(sinxcosx)
using chain rule
dy/dx=1/(sinxcosx) d/dx(sinxcosx)
dy/dx=1/(sinxcosx)((sinx(-sinx)+cosx(cosx))
dy/dx=(cos²x-sin²x)/(sinxcosx)
dy/dx=cos2x/(sinxcosx)
y=ln(ax)
Using chain rule
dy/dx=(1/ax)d/dx(ax)
dy/dx=(1/ax)a
dy/dx=1/x
or
y=ln(ax)
y=lna+lnx
dy/dx=1/x
y=2x
take log on both sides
logy=xlog2
diff both sides
dy/dx(1/y)=log2
dy/dx=ylog2
dy/dx=2xlog2
y=sinx+cosx
first derivative y’=cosx-sinx
second derivative y’’=-sinx-cosx
third derivative y’“=-cosx+sinx
fourth derivative =sinx+cosx
hence the 2016th will be same as fourth derivtive
and 2019th will be same as third derivative i.e.=sinx-cosx
Find the value of d2x/dx2 at the point defined by the given value of t.
x = 5 sin t, y = 5 cos t, t = 3π/4
x = 5 sin t, y = 5 cos t, t = 3π/4
D2Y/D2X=(D2Y/D2t/(D2x/D2t)
dy/dx=-5sint
D2Y/D2t=-5cost and
dx/dt=5cost
D2x/D2t=-5sint
D2Y/D2X=(D2Y/D2t/(D2x/D2t)=(-5cost /-5sint)
=cot(t)
=cot(3Π/4)
=-1
No comments:
Post a Comment