The derivative function
$f'(x)$ is no longer new to you: you know it is the function that gives the gradient at each point, that its sign tells you whether the function is increasing or decreasing, and that $f'(x)=0$ points to the candidates for an extremum. What is new in 2nd Bachillerato is three situations that 1st Bachillerato did not present: differentiating many times in a row and finding the pattern, differentiating a power with a variable exponent, and differentiating a curve that is not solved for $y$.
Review: the sign of $f'$ and relative extrema
Find the equation of the tangent line to the graph of the function $f(x) = x^2 - 4x + 1$ at the point with abscissa $x = 3$.
Find the equation of the tangent line to the graph of the function $f(x) = x^2 - 4x + 1$ at the point with abscissa $x = 3$.
Successive derivatives and their pattern
Given the function $f(x) = \ln(2x - 3)$, find the general expression for its nth derivative $f^{(n)}(x)$ and use it to calculate the value of the fourth derivative at…
Given the function $f(x) = \ln(2x - 3)$, find the general expression for its nth derivative $f^{(n)}(x)$ and use it to calculate the value of the fourth derivative at the point $x = 2$.
Calculate the second-order derivative (second derivative) of the function $f(x) = x^3 - 6x^2 + 9x - 2$. For which value of $x$ does this second derivative equal zero?…
Calculate the second-order derivative (second derivative) of the function $f(x) = x^3 - 6x^2 + 9x - 2$. For which value of $x$ does this second derivative equal zero? (This point is important for studying curvature, which you will see later).
Logarithmic differentiation: when the exponent is also variable
Use the technique of logarithmic differentiation to find the derivative of the following non-algebraic power function: $f(x) = (x^2 + 1)^{\cos(x)}$
Use the technique of logarithmic differentiation to find the derivative of the following non-algebraic power function:
$f(x) = (x^2 + 1)^{\cos(x)}$
Implicit differentiation
Given the equation of the curve implicitly defined by $x^3 + y^3 - 3xy = 0$ (known as the Folium of Descartes), calculate the value of the gradient of the tangent line…
Given the equation of the curve implicitly defined by $x^3 + y^3 - 3xy = 0$ (known as the Folium of Descartes), calculate the value of the gradient of the tangent line to the curve at the point $( \frac{3}{2}, \frac{3}{2} )$.