Slope of line tangent to a curve at a given point, using First Principles Find, from first principles, the gradient of the tangent to the curve $y = 5 - x^2$ at the point $(1,4)$ on the curve.
So I'm currently lost on this question can some one please show me the solutions of how to complete this question and i may use it as an guide for future question. Thanks a lot!

By first principles, perhaps you mean you need to use the definition of the derivative, $$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$ to find the slope (i.e. gradient) of the tangent line to the curve at the point $(1, 4)$, where the curve in question is given by: 
$$y = f(x) = 5 - x^2$$
$$\begin{align} f'(x) =\lim_{h \to 0} \frac{f(x + h) - f(x)}{h} & = \lim_{h \to 0} \frac{\Big(5 - (x + h)^2\Big) - (5 - x^2)}{h} \\ \\
& = \lim_{h\to 0}\frac{5 - (x^2 + 2hx + h^2) - 5 + x^2}{h} \\ \\
& = \lim_{h\to 0}\frac {-2hx - h^2}{h} \\ \\
& = \lim_{h\to 0}\frac{-h(2x + h)}{h} \\ \\
& = \lim_{h\to 0} -(2x + h)\\ \\
f'(x) & = -2x  \end{align}$$
Now, at $(1, 4), \;x = 1$, so the slope or gradient at that point is given by $f'(1) = -2(1) = -2$. 
Below, I've posted an image of the line tangent to the curve $y = 5 - x^2$ at the point $(1, 4)$, along with the the curve itself. (Image compliments of WolframAlpha).

Side note:
We can find the equation line tangent to $y = 5-x^2$ at the point $(1,4)$ knowing the slope of the line is $-2$, and the fact that $(1, 4)$ lies on that line. We know the equation of a line can be written in the form $(y - y_0) = m(x - x_0)$, where $m$ represents the slope of the line, and $(x_0, y_0)$ represents a point on the line.
In our case, we've found slope to be given by $-2$, and $(x_0, y_0) = (1, 4)$. So, $$y - 4 = -2(x - 1) \iff y - 4 = -2x + 2 \iff y = 6 - 2x$$
