Differentiate the following:
Find the derivative of the following:.
For each of the following curves and given points:
Find an expression for \dfrac{dy}{dx}.
Find the exact value of the gradient of the curve at the given point.
y = \cos 3 x at x = \dfrac{\pi}{18}.
y = \sin 4 x at x = \dfrac{\pi}{16}.
y = \sin ^{2}\left( 4 x\right) at x = \dfrac{\pi}{32}.
y = \cos ^{2}\left( 2 x\right) at x = \dfrac{\pi}{24}.
Consider the equation y = e^{\tan \left( - 6 x\right)}.
Let f \left( x \right) = \tan \left( - 6 x\right). Find f' \left( x \right).
Hence, differentiate y = e^{\tan \left( - 6 x\right)}.
Consider the expression e^{\cos x} \sin \left(e^{x}\right).
If u = e^{\cos x}, find \dfrac{d u}{d x}.
If v = \sin \left(e^{x}\right), find \dfrac{d v}{d x}.
Hence, find the derivative of y = e^{\cos x} \sin \left(e^{x}\right).
Consider the function y = \dfrac{4 x^{2} + e^{x}}{\cos 7 x}.
If u = 4 x^{2} + e^{x}, find u'.
If v = \cos 7x, find v'.
Hence, find y'.
Consider the function y = x e^{x}.
Show that e^{x + \ln x} = x e^{x}.
Hence, find \dfrac{d y}{d x}, without using the product rule.
Determine \dfrac{d y}{d x}, given that y = u^{5} and u = \ln \left(x + 5\right).
Find the derivative of the following:
y = \dfrac{x}{3} e^{ - 3 x}
y = \left( 4 x + 6\right) \tan 4 x
g \left( x \right) = \dfrac{x}{3} e^{ - 3 x} - \left( 4 x + 6\right) \tan 4 x
Consider g \left( x \right) = \sqrt { \left( e^{ 3 x} + x^{ - 4 } - \tan \dfrac{\pi}{4} x \right) } .
Find the derivative of y = e^{ 3 x} + x^{ - 4 } - \tan \dfrac{\pi}{4} x
Hence, differentiate g \left( x \right) = \sqrt{y}, expressing your answer in exact form.
Consider the expression e^{x} \left(1 + \ln 0.3 x\right)^{ - 2 }.
If y = \left(1 + \ln \left( 0.3 x\right)\right)^{ - 2 }, find \dfrac{d y}{d x}.
Hence, find the derivative of y = e^{x} \left(1 + \ln 0.3 x\right)^{ - 2 }.
Consider the equation y = \left( 2 \ln \left( - 3 x - x^{2}\right) - \cos \left( \dfrac{\pi}{3} x\right) + \dfrac{1}{x^{4}}\right)^{3}.
Find the derivative of 2 \ln \left( - 3 x - x^{2}\right).
Hence differentiate y = \left( 2 \ln \left( - 3 x - x^{2}\right) - \cos \left( \dfrac{\pi}{3} x\right) + \dfrac{1}{x^{4}}\right)^{3}.
Consider the function y = \dfrac{\sin ^{2}\left(x\right)}{\cos x}.
Prove that \dfrac{\sin ^{2}\left(x\right)}{\cos x} = \sec x - \cos x.
Differentiate f \left( x \right) = \sec x.
Hence find \dfrac{dy}{dx}, without using the product or quotient rule.
The graph of the function f \left( x \right) = e^{ 2 x} \sin 3 x is shown:
Find f' \left( x \right).
Find the x-intercepts, B and C.
Determine the coordinates of point A, correct to two decimal places.
Consider the function f \left( x \right) = x^{3} \ln x, over the domain e^{ - \frac{1}{2} } \leq x \leq e^{ - \frac{1}{4} }.
Find an expression for f' \left( x \right).
Find the exact values of x such that f' \left( x \right)=0.
Complete the table of values:
x | e^{ - \frac{1}{2} } | e^{ - \frac{1}{3} } | e^{ - \frac{1}{4} } |
---|---|---|---|
f (x) |
Hence, state the nature and location of any stationary points.
Calculate the exact global minimum over the domain.
Calculate the exact global maximum over the domain.
Consider the function y = \dfrac{\ln 3 x}{e^{ 3 x}}.
Find the x-intercept of the function.
Find an expression for \dfrac{d y}{d x}.
Show that the turning point of the graph occurs when 3 x \ln 3 x - 1 = 0.
An approximation to the solution of the equation 3 x \ln 3 x - 1 = 0 is x = 0.58.
Complete the table of values, correct to two decimal places.
x | 0.01 | 0.58 | 1.58 |
---|---|---|---|
\dfrac{dy}{dx} | 0 |
Hence, state the nature of the turning point at x = 0.58.
State the coordinates of the turning point, correct to two decimal places.
Determine the values of the non-zero constants, a and b, for the following function, given it has a turning point at \left(0.25, 1\right):
f \left( x \right) = a x e^{ b x}Consider the function f \left( x \right) = \ln \left(\sin 2 x\right).
State the values of x between 0 and 2 \pi for which the function is defined.
Determine the values of x in the domain 0 \leq x \leq 2 \pi, for which the function has a maximum value.
State the maximum value of the function.
The curve y = x \cos x passes through the point Q, \left(\dfrac{\pi}{2}, 0\right).
Find the equation of the tangent at point Q.
Find the equation of the tangent to the following curves:
y = e^{x} - 3 \sin x at x = \dfrac{3 \pi}{2}.
y = e^{\cos x} at x = \dfrac{3 \pi}{2}.
Researchers have created a model to project the country’s population for the next 10 years, where P is the population (in thousands), t years from now. The model is defined by the function: P \left( t \right) = \dfrac{57\,460 e^{\frac{t}{7}}}{t + 13}
State the current population of the country.
According to the model, state the current rate of growth of the population, to the nearest thousand.
Find the rate of population growth 7 months from now, to the nearest thousand.
Find the rate of population growth 10 years from now, to the nearest thousand.
The displacement of a particle moving in rectilinear motion is given by: x \left( t \right) = \left(t - 2\right)^{2} + \sin \left(t - 4 \pi\right) + 3 t + 25
State the initial displacement of the particle.
Write an expression for the velocity of the particle, v \left( t \right) = x' \left( t \right) .
Write an expression for the acceleration of the particle, a \left( t \right) = v' \left( t \right).
A charged particle moves back and forth about the fixed point x = 0 (called the origin). Its position, x \text{ cm} from the origin, after t seconds is given by the equation:
x = \sin \left( \pi e^{ 2 t}\right)Find the particle's initial position.
Find an expression for the velocity of the particle , v \left( t \right) = x' \left( t \right).
Find the exact times for the first and second occasion that the particle comes to a stop.
Describe the position of the particle when it first comes to a stop, v \left( t \right) = 0.
Describe the position of the particle when it comes to a stop for the second time.