Calculus - derivatives of xtan(x)

[jendoley]

Find the first and second derivative--simplify your answer.

y=x tanx

I solved the first derivative.
y'=(x)(sec^2(x)) +(tanx)(1)
y'=xsec^2(x) +tanx

I don't know about the second derivative though.

 

[Mark44]

Find the first and second derivative--simplify your answer.

y=x tanx

I solved the first derivative.
y'=(x)(sec^2(x)) +(tanx)(1)
y'=xsec^2(x) +tanx

I don't know about the second derivative though.

Take the derivative of y' to get y''. You will need the product rule and the chain rule.

 

[lurflurf]

(u v)''=u'' v+2 u' v'+u v''
recall
x''=0
and
tan'(x)=sec(x)^2=1+tan(x)^2
so
tan''(x)=(1+tan(x)^2)'=2 tan(x) tan'(x)

 

[jendoley]

That's what I don't get how to do... The second derivative. I'm assuming I have the first derivative done right. I'm lost after that.

 

[lurflurf]

To find the second derivative take derivative of te derivative.
your function is of the form
y=u v
where u=x and v=tan(x)
y'=u' v+u v'
y''=u'' v+2u' v'+u v''
now we we know u' and v' we need only find u'' and v'' and substitute them in
u=x
u'=1
u''=0
v=tan(x)
v'=1+tan(x)^2=sec(x)^2
v''=2 tan(x) tan'(x)=2 tan(x)+2 tan(x)^3=2 tan(x) sec(x)^2