Q. 1 - Calculus

Question

Determine whether each of the statements that follow is true or false. If a statement is true, explain why. If a statement is false, provide a counterexample.

Part (a): True or False: The chain rule is used to differentiate compositions of functions.

Part (b): True or False: If f and g are differentiable functions, then the derivative of $f\circ g$ is equal to the derivative of $g\circ f$. Part (c): True or False: If f and g are differentiable functions, then $\frac{d}{dx}\left(f\left(g\left(x\right)\right)\right)=f\text{'}\left(x\right)g\text{'}\left(x\right)$.

Part (d): True or False: If u and v are differentiable functions, then $\frac{d}{dx}\left(k\left(h\left(x\right)\right)\right)=k\text{'}\left(h\left(x\right)\right)h\text{'}\left(x\right)$.

Part (e): True or False: If h and k are differentiable functions, then $\frac{d}{dx}\left(k\left(h\left(x\right)\right)\right)=k\text{'}\left(h\left(x\right)\right)h\text{'}\left(x\right)$.

Part (f): True or False: If y is an implicit function of x, then there can be more than one y-value corresponding to a given x-value.

Part (g): True or False: The graph of an implicit function can have vertical tangent lines.

Part (h): True or False: If y is an implicit function of x and $\frac{d}{dx}{|}_{x=2}=0$, then the graph of the implicit function has a horizontal tangent line at $(2, 0)$.

Step-by-Step Solution

Answer

Part (a): The given statement is true.

Part (b): The given statement is false. 

Part (c): The given statement is false. 

Part (d): The given statement is false. 

Part (e): The given statement is true. 

Part (f): The given statement is true. 

Part (g): The given statement is true. 

Part (h): The given statement is false. 

Part (a) Step 1. Determine if the statement is true or false.

The chain rule is used to differentiate compositions of functions.

Consider $f\left(u\left(x\right)\right)$ is a composition of functions. Then for all values of x at which u is differentiable at x and f is differentiable at $u\left(x\right)$, the derivative of f with respect to x is equal to the product of the derivative of f  with respect to u and the derivative of u with respect to x. In prime notation, the rule is given below,

$\left(f\circ u\right)\text{'}\left(x\right)=f\text{'}\left(u\left(x\right)\right)u\text{'}\left(x\right)$

Thus, the given statement is true.

Part (b) Step 1. Determine if the statement is true or false.

If f and g are differentiable functions, then the derivative of $f\circ g$ is equal to the derivative of $g\circ f$.

Consider $f\left(u\left(x\right)\right)$ is a composition of functions. Then for all values of x at which u is differentiable at x and f is differentiable at $u\left(x\right)$, the derivative of f  with respect to x is equal to the product of the derivative of f  with respect to u and the derivative of u with respect to x. In prime notation, the rule is given below,

$\left(f\circ u\right)\text{'}\left(x\right)=f\text{'}\left(u\left(x\right)\right)u\text{'}\left(x\right)$. Then,

$$\begin{gathered} \left(g\circ f\right)\text{'}\left(x\right)=g\text{'}\left(f\left(x\right)\right)f\text{'}\left(x\right), \\ \left(f\circ g\right)\text{'}\left(x\right)=f\text{'}\left(g\left(x\right)\right)g\text{'}\left(x\right) \end{gathered}$$

Thus, the given statement is false.

Part (c) Step 1. Determine if the statement is true or false.

If f and g are differentiable functions, then,

$\frac{d}{dx}\left(f\left(g\left(x\right)\right)\right)=f\text{'}\left(x\right)g\text{'}\left(x\right)$

Consider $f\left(u\left(x\right)\right)$ is a composition of functions. Then for all values of x at which u is differentiable at x and f is differentiable at $u\left(x\right)$, the derivative of f with respect to x is equal to the product of the derivative of f with respect to u and the derivative of u with respect to x. In prime notation, the rule is given below,

$\left(f\circ u\right)\text{'}\left(x\right)=f\text{'}\left(u\left(x\right)\right)u\text{'}\left(x\right)$. Then,

$\frac{d}{dx}\left(f\left(g\left(x\right)\right)\right)=f\text{'}\left(g\left(x\right)\right)g\text{'}\left(\left(x\right)\right)$

Thus, the given statement is false.

Part (d) Step 1. Determine if the statement is true or false.

If u and v are differentiable functions, then,

$\frac{d}{dx}\left(u\left(v\left(x\right)\right)\right)=u\text{'}\left(v\text{'}\left(x\right)\right)$

Consider $f\left(u\left(x\right)\right)$ is a composition of functions. Then for all values of x at which u is differentiable at x and f is differentiable at $u\left(x\right)$, the derivative of f  with respect to x is equal to the product of the derivative of f  with respect to u and the derivative of u with respect to x. In prime notation, the rule is given below,

$\left(f\circ u\right)\text{'}\left(x\right)=f\text{'}\left(u\left(x\right)\right)u\text{'}\left(x\right)$. Then,

$\frac{d}{dx}\left(u\left(v\left(x\right)\right)\right)=u\text{'}\left(v\left(x\right)\right)v\text{'}\left(x\right)$

Thus, the given statement is false.

Part (e) Step 1. Determine if the statement is true or false.

If h and k are differentiable functions, then,

$\frac{d}{dx}\left(k\left(h\left(x\right)\right)\right)=k\text{'}\left(h\left(x\right)\right)h\text{'}\left(x\right)$

Consider $f\left(u\left(x\right)\right)$ is a composition of functions. Then for all values of x at which u is differentiable at x and f is differentiable at $u\left(x\right)$, the derivative of f  with respect to x is equal to the product of the derivative of f with respect to u and the derivative of u with respect to x. In prime notation, the rule is given below,

$\left(f\circ u\right)\text{'}\left(x\right)=f\text{'}\left(u\left(x\right)\right)u\text{'}\left(x\right)$. Then,

$\frac{d}{dx}\left(k\left(h\left(x\right)\right)\right)=k\text{'}\left(h\left(x\right)\right)h\text{'}\left(x\right)$

Thus, the given statement is true.

Part (f) Step 1. Determine if the statement is true or false.

If y is an implicit function of x, then there can be more than one y-value corresponding to a given x-value.

The equations which have more than one value of y for each x value, and which are not differentiable in simple way, those equations are called y is an implicit function of x.

Thus, the given statement is true.

Part (g) Step 1. Determine if the statement is true or false.

The graph of an implicit function can have vertical tangent lines.

The equations which have more than one value of y for each x value, and which are not differentiable in simple way, those equations are called y is an implicit function of x.

Substitute $x=0$, one can find out the vertical tangent lines of the implicit function.

Thus, the given statement is true.

Part (h) Step 1. Determine if the statement is true or false.

The equations which have more than one value of y for each x value, and which are not differentiable in simple way, those equations are called y is an implicit function of x.

Substitute $x=0$, one can find out the vertical tangent lines of the implicit function.

Thus, the given statement is false.