To determine
To find: The derivative of the function y=x+x+x.
Answer
The derivative of y=x+x+x is 1+1+12x2x+x2x+x+x_.
Explanation
Given:
The function is y=x+x+x.
Derivative rule used:
Chain Rule:
If h is differentiable at x and g is differentiable at h(x), then the composite function F=g∘h defined by F(x)=g(h(x)) is differentiable at x and F′ is given by the product
F′(x)=g′(h(x))⋅h′(x) (1)
Power Rule:
If n is positive integer, then ddx(xn)=nxn−1 (2)
Calculation:
Obtain the derivative y.
y′=ddx(y)=ddx(x+x+x)
Let h(x)=x+x+x and g(u)=u where u=h(x)
Apply the chain rule as shown in equation (1),
y′(x)=g′(h(x))⋅h′(x) (3)
The derivative g′(h(x)) is computed as follows,
g′(h(x))=g′(u)=ddu(g(u))=ddu(u)=ddu(u)12
Apply the power rule as shown in equation (2),
g′(h(x))=12(u)12−1=12(u)1−22=12(u)−12=12u
Substitute u=x+x+x in the above equation,
g′(h(x))=12x+x+x.
Thus, the derivative of g′(h(x))=12x+x+x.
The derivative of h(x) is computed as follows,
h′(x)=ddx(x+x+x)=ddx(x)+ddx(x+x)
h′(x)=1+ddx(x+x) (4)
Obtain the derivative of ddx(x+x)
Let k(x)=x+x and s(u)=u where u=k(x)
Apply the chain rule as shown in equation (1),
ddx(x+x)=s′(k(x))⋅k′(x) (5)
The derivative of s′(k(x)) computed as follows,
s′(k(x)) =s′(u)=ddu(s(u))=ddu(u)=ddu(u)12
Apply the power rule as shown in equation (2),
s′(k(x))=12(u)12−1=12(u)1−22=12(u)−12=12u
Substitute u=x+x in the above equation,
s′(k(x))=12x+x
Thus, the derivative of s′(k(x))=12x+x.
The derivative of k(x) is computed as follows,
k′(x)=ddx(x+x)=ddx(x)+ddx(x)
Apply the power rule as shown in equation (2),
k′(x)=(1x1−1)+(12x12−1)=1+12x1−22=1+12x−12=1+12x
Thus, the derivative of k′(x)=1+12x.
Substitute 12x+x for s′(k(x)) and 1+12x for k′(x) in equation (5),
ddx(x+x)=12x+x(1+12x)=1+12x2x+x
Substitute 1+12x2x+x for ddx(x+x) in equation (4),
h′(x)=1+ddx(x+x)=1+1+12x2x+x
Substitute 12x+x+x for g′(h(x)) and 1+1+12x2x+x for h′(x) in equation (3),
g′(h(x))⋅h′(x)=12x+x+x(1+1+12x2x+x)=1+1+12x2x+x2x+x+x
Therefore, The derivative of y=x+x+x is 1+1+12x2x+x2x+x+x_.