To find the derivative of the function
f(x)=‌ at
x=0, we will first find the general expression for
f′(x) using the quotient rule and the chain rule. The quotient rule states that if
f(x)=‌, then
f′(x)=‌| g′(x)h(x)−g(x)h′(x) |
| h(x)2 |
.
Let's break down the given function into
g(x)=(x+1)7√1+x2 and
h(x)=(x2−x+1)6.
We then calculate the derivatives of
g(x) and
h(x).
For
g(x), we use the product rule where if
g(x)=u(x)v(x), then
g′(x)=u′(x)v(x)+u(x)v′(x) :
Let
u(x)=(x+1)7 and
v(x)=√1+x2.
Using the power rule,
u′(x)=7(x+1)6.
For
v(x), apply the chain rule:
v′(x)=‌⋅2x=‌.
Now,
g′(x)=u′(x)v(x)+u(x)v′(x)=7(x+1)6√1+x2+(x+1)7‌.
For
h(x), use the chain rule and power rule:
First, notice
h(x)=(x2−x+1)6.
This can be found as
h′(x)=6(x2−x+1)5⋅‌(x2−x+1).
Now,
‌(x2−x+1)=2x−1.
Thus,
h′(x)=6(x2−x+1)5(2x−1).
Now, applying the quotient rule:
f′(x)=‌| g′(x)h(x)−g(x)h′(x) |
| h(x)2 |
.To calculate
f′(0), we plug
x=0 into
g′(x) and
h′(x) :
‌g′(0)=7(0+1)6√1+02+(0+1)7‌=7‌h(0)=(02−0+1)6=1‌h′(0)=6(02−0+1)5(2×0−1)=−6‌g(0)=(1)7√1=1Thus,
f′(0)=‌=7+6=13.
The value of
f′(0) is 13 . So, the correct answer is Option C: 13 .