Differential Calculus

We are given the function , where , and we are asked to find the minimum value of .

To find the minimum value, we first compute the derivative of using the product rule. The function is a product of and , so we differentiate each part:

Let and . Then, using the product rule:

,

where and .

Thus, the derivative is:

.

Now, set to find the critical points:

,

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Since , we must have ,

which simplifies to:

,

so:

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This is the critical point. To determine if this is a minimum, we compute the second derivative of :

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At , we have:

,

so:

,

which is positive, indicating that the critical point is a minimum.

Thus, the minimum value of occurs at . To find the minimum value of , substitute into the original function:

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The correct answer is .

We are given that the derivative of the function . We need to determine the interval where the function is increasing.

For a function to be increasing, its derivative must be positive. Therefore, we need to find where .

The expression for is:

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The denominator is always positive for all real values of , since for all . Therefore, the sign of is determined by the numerator, .

For , we require , which simplifies to:

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Thus, the function is increasing for .

The correct answer is .

We are given the equation of the curve and need to find the slope of the tangent line at the point .

To find the slope of the tangent line, we need to compute using implicit differentiation.

Differentiate both sides of the equation with respect to :

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Now, apply the derivative rules:

• For , .
• For , we apply the product rule: .
• For , apply the chain rule: .
• The derivative of the constant is 0.

Thus, the differentiated equation is:

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Simplify the equation:

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Now, solve for :

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Substitute and into this expression to find the slope at the point :

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The correct answer is .

We are given that the tangent line to the graph of the function at the point has an x-intercept of and a y-intercept of -10. We are asked to find .

Let the equation of the tangent line be of the form , where is the slope of the tangent line.

We are given the x-intercept and y-intercept of the tangent line:

• The x-intercept is the point where , so we have .
• The y-intercept is the point where , so we have .

First, solve the equation for the x-intercept:

,

,

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Next, solve the equation for the y-intercept:

,

,

Substitute into this equation:

,

Multiply through by 2 to eliminate the fraction:

,

,

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Thus, .

The correct answer is 5.

We are given the function , and we need to find .

To solve this, we will use the chain rule for differentiation. Let:

, so that .

Now, we differentiate with respect to using the chain rule:

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Next, we need to compute :

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Thus, the derivative of is:

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Now, we substitute into this expression:

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Since and , we have:

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.

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The correct answer is 5(3 - π).

We are given that for . We need to find the function such that the derivative is given by:

We will begin by differentiating using the chain rule and logarithmic differentiation.

Let , and take the natural logarithm on both sides:

Now, differentiate both sides with respect to :

On the left-hand side, we use the chain rule:

Now, differentiate the right-hand side using the product rule:

Next, differentiate with respect to :

Thus, the derivative becomes:

Comparing this with the given derivative, we can see that:

Therefore, the correct answer is:

sin x sec² x

We are given that , and we need to find the value of such that , where and are constants.

First, simplify the expression for :

Now, differentiate with respect to :

Thus, we have , which is of the form , where and .

Therefore, the value of is:

Step 1: Differentiate the function with respect to .

Step 2: Differentiate the first term using the product rule: This can be rewritten as:

Step 3: Differentiate the second term :

Step 4: Combine the derivatives: Or equivalently:

Conclusion: The correct derivative matches option (E):