#### To determine

**To find:** The value h′(1).

#### Answer

The value h′(1) is h′(1)=65_.

#### Explanation

**Given:**

The function is h(x)=4+3f(x), where f(1)=7,f′(1)=4.

**Result used: The Power Rule combined with the Chain Rule**

If *n* is any real number and g(x) is differentiable function, then

ddx[g(x)]n=n[g(x)]n−1g′(x) (1)

**Calculation:**

Obtain the derivative of h(x)=4+3f(x)*.*

h′(x)=ddx(h(x))=ddx(4+3f(x))=ddx[4+3f(x)]12

Apply the power rule combined with the chain rule as shown in equation (1),

h′(x)=12[4+3f(x)]12−1ddx(4+3f(x))=12[4+3f(x)]1−22(ddx(4)+ddx(3f(x)))=12[4+3f(x)]−12(0+(3f′(x)))

h′(x) =3f′(x)24+3f(x) (2)

Substitute x=1 in equation (2),

h′(1) =3f′(1)24+3f(1)

Substitute f(1)=7 and f′(1)=4 in the above equation,

h′(1) =3(4)24+3(7)=1224+21=12225=122⋅5

Simplify further and obtain the values of h′(1)*.*

h′(1)=1210=65

Therefore, the derivative of h(1)=4+3f(1) is h′(1)=65_.