Yes, this is a homework problem :), so I'm not looking for the answer just how to approach it.Find derivative using a table of values
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$\begingroup$$$h(x)=xf(x)+2g(x)$$
$$\frac{d}{dx}h(x)=\frac{d}{dx}(xf(x) + 2g(x))$$
Distributing the derivative we get this main equation:
$$\frac{d}{dx}h(x)=\frac{d}{dx}(xf(x))+\frac{d}{dx}(2g(x)) $$
$$\frac{d}{dx}(xf(x))=f(x)+x\frac{d}{dx}f(x)=f(x)+xf'(x)$$
Plug this into the main equation and simplify, to get:
$$\frac{d}{dx}h(x)=[f(x)+xf'(x)]+[2g'(x)] $$
Now you can find all the values on the right hand side of the above equation in the table give. So just substitute to get the value of $h'(x)$.
$\endgroup$ 4 $\begingroup$Note:
- $\frac{dh}{dx}(a)=\left.\frac{dh}{dx}\right|_{x=a}$ i.e. derivative evaluated at given point $a$ (differentiating with respect to $x$and then evaluating the function at $a$.
Product rule of differentiation:
$\frac{d}{dx} (u\cdot v) = v \cdot \frac{du}{dx} + u \cdot \frac{dv}{dx}$
Only these two concepts are required to solve your problem. Apply the product rule to differentiate and then evaluate the function obtained at the given point.
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