How solve it? Topic: DERIVATE

Please I need help I do not know how to solveenter image source here

2 Answers
Jun 1, 2018

#f(x)=sqrt(2x)#

First, let's rewrite the square root as a #1//2# power.

#f(x)=(2x)^(1/2)#

Now, we need to recognize that these can be split up as a constant and a variable function.

#f(x)=2^(1/2)*x^(1/2)#

When we differentiate, multiplicative constants like the #2^(1/2)# here simply stay on the "outside," that is, we don't do anything to them.

To differentiate #x^(1/2)#, we use the power rule, which says that #d/dxx^n=nx^(n-1)#.

Then, we see that:

#f'(x)=2^(1/2)*(1/2x^(1/2-1))#

Now simplifying:

#f'(x)=2^(1/2)/2x^(-1/2)#

#f'(x)=1/(2^(1/2)*x^(1/2))#

#f'(x)=1/sqrt(2x)#

So at #x=5/3#, the derivative is equal to:

#f'(5/3)=1/sqrt(2*5/3)=1/sqrt(10/3)=sqrt(3/10)#

Jun 1, 2018

#f'(x)=2/[2sqrt(2x)]=1/sqrt(2x)#
#f'(5/3)=1/sqrt(10/3)=sqrt3/sqrt10#

Explanation:

show below:

#f(x)=sqrt(2x)#

#f'(x)=2/[2sqrt(2x)]=1/sqrt(2x)#

The derivative at #x=5/3# equal

#f'(5/3)=1/sqrt(10/3)=sqrt3/sqrt10#

#"Note that"#

#color(red)[y=sqrtx]#

#color(red)[y'=1/[2sqrtx]*x']#

#color(red)[sqrt[a/b]=sqrta/sqrtb]#