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Calculus
Section 2.8
With question 32 we end Section 2.8 and Chapter 2.
Section 2.8
With question 32 we end Section 2.8 and Chapter 2.
(d/dx)(1/(1 + sqrt(x))). ......aply exponent rule 1/a=a^-1
=(d/dx)((1 + sqrt(x))^-1)..............apply chain rule
=-1/(1 + sqrt(x))^2*(d/dx)((1 + sqrt(x))............(d/dx)((1 + sqrt(x))=1/(2sqrt(x))
=-1/(1 + sqrt(x))^2*1/(2sqrt(x)).......simplify
=-1/((1 + sqrt(x))^2*2sqrt(x))
so,
View attachment 3499
correct
yes -> The range is the set of possible output values, which are shown on the y-axis.
f(x)=1/(1+sqrt(x))
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range: { y element R : 0<y<=1}
View attachment 3517
f'(x)=-1/((1 + sqrt(x))^2*2sqrt(x))
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range:{ y element R : y<0} (all negative real numbers)
View attachment 3518
finding the range for Definition of Derivatives????? what that means
you can find the derivative of the function using the definition of derivative