The derivative $\mathrm{f}^{\prime}(x)$ gives the tangent gradient where the curve is differentiab

Question

The derivative $\mathrm{f}^{\prime}(x)$ gives the tangent gradient where the curve is differentiable. $\dfrac{\mathrm{d}y}{\mathrm{d}x}=\mathrm{f}^{\prime}(x)\qquad\dfrac{\mathrm{d}}{\mathrm{d}x}(x^n)=nx^{n-1}$ Apply the power rule where the function and derivative exist. Rewrite roots and reciprocals as powers first. For a straight line, the gradient is $\dfrac{y_2-y_1}{x_2-x_1}$: vertical change divided by horizontal change. Example For $y=3$, the gradient is $0$. For both $y=2x$ and $y=2x-3$, the gradient is $2$. For $y=x^2$, the gradient depends on the point on the curve. $\dfrac{\mathrm{d}y}{\mathrm{d}x}$${}=2x$ At $x=-1,0,1,2$, the gradients are respectively $-2,0,2,4$. At $x=a$, the gradient is $2a$.

Accepted Answer

$y=3\mapsto0$, $y=2x\mapsto2$, $y=2x-3\mapsto2$, $y=x^2\mapsto2x$

Derivative as a Gradient Function

Derivative as a Gradient Function

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