What is the derivative of #sinx^tanx#?
↳Redirected from
"How many antibonding orbitals are there in Benzene?"
#cos(x^(tan(x)))x^(tan(x))(tan(x)/x+ln(x)*sec^2(x))#
[Assuming you meant #sin(x^(tan(x)))#]
We have:
#d/dx[sin(x^(tan(x)))]# We use the chain rule:
#d/dx[g(h(x))]=g'(h(x))*h'(x)#
Also remember that #d/dx[sin(x)]=cos(x)#
#=>cos(x^(tan(x)))*d/dx[x^(tan(x))]#
Here is a rule that you may not be familiar with:
#d/dx[(f(x))^(g(x))]=(f(x))^(g(x))*d/dx[ln(f(x))*g(x)]#
#=>cos(x^(tan(x)))*x^(tan(x))*d/dx[ln(x)*tan(x)]#
Use the product rule:
#d/dx(f(x)*g(x))=f'(x)*g(x)+f(x)*g'(x)#
#=>cos(x^(tan(x)))x^(tan(x))(d/dx[ln(x)]*tan(x)+ln(x)*d/dx(tan(x)))#
Remember that:
#d/dx(ln(x))=1/x#
#d/dx(tan(x))=sec^2(x)#
#=>cos(x^(tan(x)))x^(tan(x))(1/x*tan(x)+ln(x)*sec^2(x))#
#=>cos(x^(tan(x)))x^(tan(x))(tan(x)/x+ln(x)*sec^2(x))#
#(sinx)^(tanx)*(((cosx)(tanx))/(sinx)+ln(sinx)*sec^2x)#
[Assuming you meant #(sinx)^(tanx)#]
We have:
#d/dx[(sinx)^(tanx)]#
We use a rule you may be unfamiliar with:
#d/dx[(f(x))^(g(x))]=(f(x))^(g(x))*d/dx[ln(f(x))*g(x)]#
Therefore, we have:
#(sinx)^(tanx)*d/dx[ln(sinx)*tanx]#
The product rule:
#d/dx(f(x)*g(x))=f'(x)*g(x)+f(x)*g'(x)#
#=>(sinx)^(tanx)*(d/dx[ln(sinx)]*tanx+ln(sinx)*d/dx[tanx])#
Remember that:
#d/dx[f(g(x))]=f'(g(x))*g'(x)#
#d/dx(lnx)=1/x#
#d/dx(tanx)=sec^2x#
#=>(sinx)^(tanx)*(1/(sinx)*d/dx(sinx)*tanx+ln(sinx)*sec^2x)#
Another thing to remember here:
#d/dx(sinx)=cosx#
#=>(sinx)^(tanx)*(1/(sinx)*cosx*tanx+ln(sinx)*sec^2x)# Simplify.
#=>(sinx)^(tanx)*(((cosx)(tanx))/(sinx)+ln(sinx)*sec^2x)#
This is our answer!
