Given that y = u^{3} and u = 2 x + 3, define y in terms of x.
Redefine the following functions as composite functions f \left( u \right) and u \left( x \right), where u \left( x \right) is a polynomial:
f \left( x \right) = \left( 5 x^{3} - 4 x^{2} + 3 x - 5\right)^{7}
f \left( x \right) = \sqrt[4]{ 2 x^{2} + 2 x + 3}
f \left( x \right) = \dfrac{1}{\left( 4 x^{2} - 3 x + 5\right)^{3}}
If u = 8 x^{2} + 3, write an expression for the following functions in the form f \left( u \right) = a u^{n}:
f \left( x \right) = 2 \left( 8 x^{2} + 3\right)^{5}
f \left( x \right) = \sqrt{ 8 x^{2} + 3}
f \left( x \right) = \dfrac{5}{8 x^{2} + 3}
f \left( x \right) = \dfrac{- 1}{\left( 8 x^{2} + 3\right)^{5}}
f \left( x \right) = - \dfrac{5}{3 \sqrt{ 8 x^{2} + 3}}
If u = e^{x} + 4 x^{2}, write expressions for the following functions in terms of u:
f \left( x \right) = 3 \left(e^{x} + 4 x^{2}\right)^{5}
f \left( x \right) = 3 \cos \left(e^{x} + 4 x^{2}\right)
f \left( x \right) = 8^{e^{x} + 4 x^{2}}
f \left( x \right) = \sin \left( 8 x^{2} + 2 e^{x}\right) + e^{x} + 4 x^{2}
To differentiate the function y = \left( 2 x^{4} + 6\right)^{5} using the substitution u = 2 x^{4} + 6:
Find \dfrac{d u}{d x}.
Express y as a function of u.
Find \dfrac{d y}{d u}.
Hence, find \dfrac{d y}{d x}.
For each of the following:
Find \dfrac{dy}{du}.
Find \dfrac{du}{dx}.
Hence, find \dfrac{dy}{dx}.
y = \left(x + 5\right)^{5}, where y = u^{5} and u = x + 5.
y = \left( 4 x + 3\right)^{ - 1 }, where y = u^{ - 1 } and u = 4 x + 3.
y = \sqrt{5 + x^{2}}, where y = \sqrt{u} and u = 5 + x^{2}.
y = \sqrt[3]{\dfrac{15}{x}}, where y = \sqrt[3]{u} and u = \dfrac{15}{x}.
Consider the function y = \left( 5 x - 7\right)^{2}.
Differentiate y by expanding the brackets first.
Differentiate y by using the chain rule.
Find the derivative of y = \sqrt{ 8 x + 5} using the chain rule. Give your answer in surd form.
Differentiate the following using the chain rule:
y = \left( 4 x + 3\right)^{9}
y = \left( 2 t^{7} + 8 t^{3} + 3 t + 5\right)^{ - 4 }
y = \left(x^{2} + x^{ - 3 }\right)^{3}
y = \sqrt[3]{x^{2} - 5 x}
y = \left( 3 x^{2} - 4 x + 2\right)^{4}
y = - 3 \left( 3 x + 4\right)^{10}
y = \dfrac{\left( 9 x + 7\right)^{\frac{4}{3}}}{3}
y = \dfrac{1}{\left(x + 6\right)^{5}}
y = \dfrac{1}{x^{4} - 4 x^{3} + 5 x}
y = \dfrac{2}{\sqrt{1 + x}}
y = \sqrt[5]{\left( 4 x + 1\right)^{6}}
y = - 4 \left( \dfrac{1}{3} x + 1\right)^{ - 6 }
y = 5 \sqrt{4 - \dfrac{1}{3} x}
y = \dfrac{2}{1 - x \sqrt{5}}
y = \left(\sqrt{x} + \dfrac{1}{\sqrt{x}}\right)^{8}
y = - 3 \left(x + \dfrac{1}{x}\right)^{5}
Find the gradient of f \left( x \right) = \left(x - 5\right)^{3} at the point \left(8, 27\right).
Find the x-coordinate of the point at which f \left( x \right) = \left(x - 2\right)^{2} has a gradient of 6.
Find the x-coordinate(s) of the point(s) at which f \left( x \right) = \left(x + 2\right)^{3} has a gradient of 48.
Consider the function g \left( x \right) = \left(3 - x^{5}\right)^{4}. Use the chain rule to evaluate g' \left( - 1 \right).
Find the values of x where the tangent of y = \left(x^{2} - 1\right)^{3} is horizontal.
Consider the function y = \sqrt{3 - 2 x}. Does there exist a point on the function that would have a horizontal tangent? Explain your answer.
Find the values of x where the derivative of y = \left( 2 x + x^{2}\right)^{5} is equal to zero.
Consider the semicircle defined as y = \sqrt{169 - x^{2}}.
Find the derivative of the function.
Sketch the graph of the semicircle.
Find the equation of the tangent at the point \left(5, 12\right).
Find the gradient of the tangent to the curve y = \left( 3 x - 1\right)^{3} at the point \left(1, 8\right).
Find f' \left( 2 \right) for the function f \left( x \right) = 2 \left(x^{2} - 7\right)^{5}.
Consider the function y = \left(x - 5\right)^{5} + 4. Determine the value(s) of x for which the tangent to the function is parallel to the x-axis.
Consider the function f \left( t \right) = \dfrac{3}{6 - t^{2}}.
Determine an expression for f' \left( t \right), expressing the derivative in positive index form.
Hence determine the gradient of the curve f \left( t \right) where t = - 2.
The curve y = \sqrt{x - 3} has a tangent with a gradient of \dfrac{1}{2} at the point P. Find the coordinates of P.
For the function f \left( x \right) = \dfrac{1}{2 x - 7}, find the values of x where f' \left( x \right) = - \dfrac{2}{25}.
Find the equation of the tangent to y = \left( 2 x + 1\right)^{4} at the point where x = - 1.
Find the equation of the tangent to y = \left( 2 x - 1\right)^{8} at the point where x = 1.