Answer:
[tex]y = x \ln 2 +\left(\ln 2 \right)^2-2 \ln 2[/tex]
Step-by-step explanation:
Differentiation is an algebraic process that finds the gradient of a curve.
At a point, the gradient of a curve is the same as the gradient of the tangent line to the curve at that point.
[tex]\boxed{\begin{minipage}{5.4 cm}\underline{Chain Rule for Differentiation}\\\\If $y=f(u)$ and $u=g(x)$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=\dfrac{\text{d}y}{\text{d}u}\times\dfrac{\text{d}u}{\text{d}x}$\\\end{minipage}}[/tex]
[tex]\boxed{\begin{minipage}{4.5 cm}\underline{Differentiating $x^n$}\\\\If $y=x^n$, then $\dfrac{\text{d}y}{\text{d}x}=nx^{n-1}$\\\end{minipage}}[/tex]
[tex]\boxed{\begin{minipage}{4.5 cm}\underline{Differentiating $\ln x$}\\\\If $y=\ln x$, then $\dfrac{\text{d}y}{\text{d}x}=\dfrac{1}{x}$\\\end{minipage}}[/tex]
Differentiate the given function using the chain rule:
[tex]\begin{aligned}f(x) & = \left(\ln x\right)^2\\\implies f'(x)& = 2\left(\ln x\right)^{2-1} \cdot \dfrac{\text{d}}{\text{d}x} \ln x\\& = 2\left(\ln x\right)^{1} \cdot \dfrac{1}{x}\\& = \dfrac{2}{x} \ln x \end{aligned}[/tex]
To find the gradient of the function at x = 2, substitute x = 2 into the differentiated function:
[tex]\implies f'(2) = \dfrac{2}{2} \ln 2 = \ln 2[/tex]
Therefore, the gradient of the function at x = 2 is ln(2).
Substitute x = 2 into the function to find the y-value of the point on the curve when x = 2:
[tex]\implies f(2)= \left( \ln 2\right)^2[/tex]
Slope-intercept form of a linear equation:
[tex]y=mx+b[/tex]
where:
Substitute the point (2, (ln 2)²) and the found gradient into the slope-intercept formula and solve for b:
[tex]\begin{aligned} y & = mx+b\\\implies \left(\ln 2 \right)^2 & = \ln 2 \cdot 2 + b\\\left(\ln 2 \right)^2 & =2\ln 2 +b\\b & = \left(\ln 2 \right)^2-2 \ln 2\end{aligned}[/tex]
Therefore, the tangent has the equation:
[tex]y = x \ln 2 +\left(\ln 2 \right)^2-2 \ln 2[/tex]
