Problem 47 - A Textbook of Differential Calculus

Question

Let \(h(x)\) be differentiable for all \(x\) and let \(f(x)=\left(k x+e^{x}\right) h(x)\) where \(k\) is some constant. If \(h(0)=5, h^{\prime}(0)=-2\) and \(f^{\prime}(0)=18\), then the value of \(k\) is equal to $\ldots \ld

Step-by-Step Solution

Answer

The value of \(k\) is 3.

Step 1: Differentiate the Product Function

To find the derivative of the function \(f(x) = (kx + e^x)h(x)\), we apply the product rule: \((uv)' = u'v + uv'\), where \(u = kx + e^x\) and \(v = h(x)\). Differentiating, we get:\[f'(x) = (k + e^x)h(x) + (kx + e^x)h'(x)\]

Step 2: Substitute Known Values at \(x = 0\)

We know that \(h(0) = 5\) and \(h'(0) = -2\). Substitute \(x = 0\) into the derivative equation:\[f'(0) = (k + e^0)h(0) + (0 + e^0)h'(0)\]Simplify this to:\[18 = (k + 1) \times 5 + 1 \times (-2)\]

Step 3: Solve for \(k\)

Now, solve \[18 = (k + 1) \times 5 - 2\]. Simplifying this, we get:\[18 = 5k + 5 - 2\]\[18 = 5k + 3\]Subtract 3 from both sides:\[15 = 5k\]Divide both sides by 5:\[k = 3\]

Key Concepts

Product Rule

The product rule is a fundamental technique in differential calculus used when differentiating the product of two functions. If you have two functions, say \(u(x)\) and \(v(x)\), their derivative when multiplied together isn't as simple as differentiating each one individually. Instead, you must apply the product rule, which states:

• \((uv)' = u'v + uv'\)

This means you take the derivative of the first function \(u\) and multiply it by the second function \(v\), then add the product of \(u\) and the derivative of \(v\).

In our problem, we applied the product rule to \(f(x) = (kx + e^x)h(x)\). Here, \(u = kx + e^x\) and \(v = h(x)\). It helps simplify the differentiation process when dealing with products, especially when functions are considered as being multiplied together.

Differentiation

Differentiation is the process of finding a derivative, which essentially indicates how a function changes as its input changes. This rate of change is vital in calculus as it provides insights into the behavior of functions.

When we differentiate, we use specific rules depending on the form of the function:

• Power Rule: \(\frac{d}{dx}[x^n] = nx^{n-1}\)
• Exponential Rule: \(\frac{d}{dx}[e^x] = e^x\)
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Differentiation is extensively used in our problem to determine the derivative of the function \(f(x)\) through the product rule methodology.

It is important to remember that the differentiation process varies depending on the function types involved, as seen with the exponential part \(e^x\) within \(f(x)\).

Function Derivative

A function derivative represents the rate at which a function's output changes with respect to changes in its input. It is expressed as \(f'(x)\) if \(f(x)\) denotes the original function.

In our exercise, the function \(f(x) = (kx + e^x)h(x)\) required us to find \(f'(x)\), its derivative. The derivative contained terms corresponding to the derivatives of each part of the product \((kx + e^x)\) and \(h(x)\), as applied in the product rule.

Finding the derivative of a function is crucial for solving many calculus problems, as it provides key insights into the slope of a tangent line to the function at any point, which signifies the function's rate of change at that point.