What is #f'(-pi/3)# when you are given #f(x)=sin^7(x)#?
↳Redirected from
"How does a vector quantity differ from a scalar quantity?"
It is #(7sqrt3)/2^7=(7sqrt3)/128#
Method
#f(x)=sin^7(x)#
It is very useful to re-write this as #f(x)=(sin(x))^7# because this makes it clear that what we have is a #7^(th)# power function.
Use the power rule and the chain rule (This combination is often called the generalized power rule.)
For #f(x)=(g(x))^n#, the derivative is #f'(x)=n(g(x))^(n-1)*g'(x)#,
In other notation #d/(dx)(u^n)=n u^(n-1) (du)/(dx)#
In either case, for your question #f'(x)=7(sin(x))^6*cos(x)#
You could write #f'(x)=7sin^6(x) *cos (x)#
At #x=- pi/3#, we have
#f'(- pi/3)=7sin^6(- pi/3) *cos (- pi/3)=7(1/2)^6(sqrt3/2)=(7sqrt3)/2^7#
#"let " y= f(x) # #=> dy/dx = f'(x)#
#=>y = sin^7(x)#
#"let " u = sin(x) => y = u^7#
#du/dx = cos(x)#
# dy/du = 7*u^6#
Now, #f'(x) = (dy)/(dx)#
#= (dy)/(du)*(du)/(dx)# {Do you agree?}
#= 7u^6*cosx#
but remember #u = sin(x)#
#=> f'(x) = 7sin^6(x)cos(x)#
#=> f'(-pi/3) = 7*(sin(-pi/3))^6**cos(-pi/3)#
# = 7(-sqrt(3)/2)^6**(1/2)#
You Have The Honor To Simplify
NOTE:
{
wondering why im doing all this "let stuff" ?
the reason is there are more than one function in #f(x)#
** there's : #sin^7(x)# and there's #sin(x) #!!
so to find the #f'(x)# i need to find the #f'# of #sin^7(x)#
AND the #f'# of #sin(x)#
that's why i need to let # y= f(x)#
then let #u = sin(x)#
}
