Question Statement

Find the slope of the tangent at $x=1$ for the following functions:

(a) $y=x^2+3x$

(b) $y=x^4-x^2$

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Background and Explanation

The slope of the tangent line to a curve at a specific point is given by the derivative of the function evaluated at that point. To find $\frac{dy}{dx}$, we apply the power rule for differentiation: $\frac{d}{dx}(x^n)=n{x}^{n-1}$. Once we have the derivative, we substitute $x=1$ to determine the instantaneous rate of change at that point.

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Solution

Part (a): $y=x^2+3x$

Given the function: $y=x^2+3x$

Differentiate with respect to $x$: $\frac{dy}{dx}=\frac{d}{dx}\left(x^2\right)+3\frac{d}{dx}(x)$

$=2x+3\cdot (1)$

$\frac{dy}{dx}=2x+3$

At $x=1$: $\frac{dy}{dx}=2(1)+3=5$

Therefore, the slope of the tangent at $x=1$ is $5$.

Part (b): $y=x^4-x^2$

Given the function: $y=x^4-x^2$

Differentiate with respect to $x$: $\frac{dy}{dx}=\frac{d}{dx}\left(x^4\right)-\frac{d}{dx}\left(x^2\right)$

$\frac{dy}{dx}=4x^3-2x$

At $x=1$: $\frac{dy}{dx}=4(1{)}^3-2(1)$

$=4(1)-2$

$=4-2$

$\frac{dy}{dx}=2$

Therefore, the slope of the tangent at $x=1$ is $2$.

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Key Formulas or Methods Used

• Power Rule: $\frac{d}{dx}(x^n)=n{x}^{n-1}$
• Sum/Difference Rule: $\frac{d}{dx}[f(x)\pm g(x)]=f^{\prime }(x)\pm g^{\prime }(x)$
• Constant Multiple Rule: $\frac{d}{dx}[c\cdot f(x)]=c\cdot f^{\prime }(x)$
• Slope of Tangent: $m={\frac{dy}{dx}|}_{x=a}$

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Summary of Steps

1. Write down the given function $y=f(x)$.
2. Differentiate with respect to $x$ using the power rule to obtain $\frac{dy}{dx}$.
3. Substitute $x=1$ into the derivative expression.
4. Simplify the arithmetic to calculate the final numerical value.
5. State the result as the slope of the tangent line at $x=1$.