Mastering Differentiation A Comprehensive Guide To Finding Derivatives

This comprehensive guide delves into the realm of differential calculus, providing step-by-step solutions and explanations for finding the derivative, denoted as ${\frac{dy}{dx}}$, of various functions and expressions. Mastering differentiation is crucial for understanding rates of change and optimization problems in mathematics, physics, engineering, and economics. This guide will equip you with the knowledge and skills to confidently tackle a wide range of differentiation challenges. Let's embark on this journey of mathematical exploration!

1. Derivative of ${ y = \sqrt{x} }$

To find the derivative of ${ y = \sqrt{x} }$, we'll employ the power rule, a fundamental concept in differential calculus. The power rule states that if ${ y = x^n }$, then ${\frac{dy}{dx} = nx^{n-1}}$. Our first key step is to rewrite the square root function as a power function. Recognizing that ${\sqrt{x}}$ is equivalent to ${x^{\frac{1}{2}}}$ allows us to directly apply the power rule. The power rule application involves bringing the exponent down as a coefficient and then reducing the exponent by 1. This process transforms ${x^{\frac{1}{2}}}$ into ${\frac{1}{2}x^{\frac{1}{2} - 1}}$. Simplifying the exponent, we get ${\frac{1}{2}x^{-\frac{1}{2}}}$ which can be further rewritten using the properties of exponents as ${\frac{1}{2\sqrt{x}}}$. This final form provides a clear and concise representation of the derivative. The derivative, ${\frac{dy}{dx}}$, represents the instantaneous rate of change of the function ${y}$ with respect to ${x}$. In the context of ${y = \sqrt{x}}$, the derivative ${\frac{1}{2\sqrt{x}}}$ tells us how the value of ${y}$ changes as ${x}$ changes. For instance, at ${x = 1}$, the derivative is ${\frac{1}{2}}$, indicating that for a small change in ${x}$ around 1, the change in ${y}$ is approximately half the change in ${x}$. As ${x}$ increases, the derivative decreases, implying that the rate of change of ${y}$ with respect to ${x}$ slows down. This behavior aligns with the shape of the square root function, which grows less steeply as ${x}$ increases. Understanding the derivative not only provides a formula for calculation but also offers insights into the behavior of the original function. This conceptual understanding is paramount in applying calculus to real-world problems. In summary, applying the power rule to ${y = \sqrt{x}}$ involves rewriting the function as a power of ${x}$, applying the power rule formula, and simplifying the result to obtain the derivative (\frac{1}{2\sqrt{x}}. This derivative quantifies the rate of change of the square root function and provides valuable information about its behavior.

Solution:

${ y = \sqrt{x} = x^{\frac{1}{2}} }$

Applying the power rule:

${ \frac{dy}{dx} = \frac{1}{2}x^{\frac{1}{2} - 1} = \frac{1}{2}x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}} }$

2. Derivative of ${ y = \cos(\sqrt{q}) }$

To determine ${\frac{dy}{dq}}$ for the function ${ y = \cos(\sqrt{q}) }$, we must employ the chain rule. The chain rule is a fundamental principle in calculus that allows us to differentiate composite functions, which are functions within functions. In this case, we have the cosine function acting on the square root of ${q}$, making it a clear candidate for the chain rule. The core idea of the chain rule is to differentiate the outer function while keeping the inner function intact, then multiply by the derivative of the inner function. This step-by-step process ensures that we account for the contributions of each function to the overall rate of change. First, we differentiate the outer cosine function. Recall that the derivative of ${\cos(u)}$ is ${-\sin(u)}$, where ${u}$ represents the inner function. Applying this, we get ${-\sin(\sqrt{q})}$. Next, we differentiate the inner function, which is ${\sqrt{q}}$. As we saw in the previous problem, the derivative of ${\sqrt{q}}$ (or ${q^{\frac{1}{2}}}$) with respect to ${q}$ is ${\frac{1}{2\sqrt{q}}}$. Now, we apply the chain rule by multiplying the derivative of the outer function by the derivative of the inner function. This gives us ${-\sin(\sqrt{q}) \cdot \frac{1}{2\sqrt{q}}}$ which simplifies to ${-\frac{\sin(\sqrt{q})}{2\sqrt{q}}}$. This final expression represents the derivative of ${y}$ with respect to ${q}$, or ${\frac{dy}{dq}}$. The significance of the chain rule lies in its ability to break down complex differentiation problems into manageable steps. By systematically differentiating the outer and inner functions and then combining the results, we can accurately find the derivative of composite functions. Without the chain rule, differentiating such functions would be significantly more challenging. Understanding the chain rule not only allows us to calculate derivatives but also provides insights into how the rates of change of different parts of a composite function interact. This holistic view is essential for applying calculus in various contexts. In essence, differentiating ${ y = \cos(\sqrt{q}) }$ involves identifying the composite structure, applying the chain rule by differentiating the outer function and then the inner function, and multiplying the results to obtain the derivative ${-\frac{\sin(\sqrt{q})}{2\sqrt{q}}}$.

Solution:

Let ${ y = \cos(\sqrt{q}) }$

Applying the chain rule:

${ \frac{dy}{dq} = -\sin(\sqrt{q}) \cdot \frac{d}{dq}(\sqrt{q}) = -\sin(\sqrt{q}) \cdot \frac{1}{2\sqrt{q}} = -\frac{\sin(\sqrt{q})}{2\sqrt{q}} }$

3. Derivative of ${ \frac{x \sin x}{\cos x} = \frac{\sin(x) \cos(x) + x}{\cos^2 x} }$

To find the derivative of the given expression, we'll utilize a combination of the quotient rule and product rule, which are essential techniques for differentiating complex functions. The given equation ${\frac{x \sin x}{\cos x} = \frac{\sin(x) \cos(x) + x}{\cos^2 x}}$ appears to be an attempt to simplify the derivative, but to verify this, we will independently find the derivative of the left-hand side and compare it to the right-hand side. First, let's consider the left-hand side, ${y = \frac{x \sin x}{\cos x}}$. This expression is a quotient of two functions, where the numerator is ${x \sin x}$ and the denominator is ${\cos x}$. The quotient rule states that if ${y = \frac{u}{v}}$, then ${\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}}$. Here, ${u = x \sin x}$ and ${v = \cos x}$. To find ${\frac{du}{dx}}$, we need to apply the product rule because ${u}$ is a product of two functions, ${x}$ and ${\sin x}$. The product rule states that if ${u = fg}$, then ${\frac{du}{dx} = f\frac{dg}{dx} + g\frac{df}{dx}}$. Applying the product rule to ${u = x \sin x}$, we get ${\frac{du}{dx} = x \cos x + \sin x}$. Next, we find ${\frac{dv}{dx}}$, which is the derivative of ${\cos x}$, which is ${-\sin x}$. Now, we can apply the quotient rule: ${ \frac{dy}{dx} = \frac{\cos x (x \cos x + \sin x) - x \sin x (-\sin x)}{\cos^2 x} }$ Simplifying this expression, we get: ${ \frac{dy}{dx} = \frac{x \cos^2 x + \sin x \cos x + x \sin^2 x}{\cos^2 x} }$ We can factor out an ${x}$ from the first and third terms in the numerator:

${ \frac{dy}{dx} = \frac{x(\cos^2 x + \sin^2 x) + \sin x \cos x}{\cos^2 x} }$ Since ${\cos^2 x + \sin^2 x = 1}$, the expression simplifies to:

${ \frac{dy}{dx} = \frac{x + \sin x \cos x}{\cos^2 x} }$ This result matches the right-hand side of the given equation. Therefore, the given equation represents the correct derivative of ${\frac{x \sin x}{\cos x}}$. In summary, to find the derivative, we applied the quotient rule along with the product rule. We systematically found the derivatives of the numerator and denominator and then combined them using the quotient rule formula. The simplification using the trigonometric identity ${\cos^2 x + \sin^2 x = 1}$ led us to the final derivative, confirming the provided equation.

Solution:

Let ${ y = \frac{x \sin x}{\cos x} }$

Applying the quotient rule:

${ \frac{dy}{dx} = \frac{\cos x \frac{d}{dx}(x \sin x) - x \sin x \frac{d}{dx}(\cos x)}{\cos^2 x} }$ Applying the product rule to ${\frac{d}{dx}(x \sin x)}$:

${ \frac{d}{dx}(x \sin x) = x \cos x + \sin x }$ Substituting back into the quotient rule:

${ \frac{dy}{dx} = \frac{\cos x (x \cos x + \sin x) - x \sin x (-\sin x)}{\cos^2 x} }$ Simplifying:

${ \frac{dy}{dx} = \frac{x \cos^2 x + \sin x \cos x + x \sin^2 x}{\cos^2 x} = \frac{x(\cos^2 x + \sin^2 x) + \sin x \cos x}{\cos^2 x} = \frac{x + \sin x \cos x}{\cos^2 x} }$

4. Derivative of ${ x^2 + \sin(x)^{\cos(x)} }$

To find the derivative of ${ y = x^2 + \sin(x)^{\cos(x)} }$, we need to tackle each term separately. The first term, ${x^2}$, is a simple power function, and its derivative can be found using the power rule. The second term, ${\sin(x)^{\cos(x)}}$, is more complex as it involves a variable raised to a variable power, which requires logarithmic differentiation. This process is essential for handling functions of the form ${f(x)^{g(x)}}$. Let's first address the derivative of ${x^2}$. Applying the power rule, we get ${2x}$. Now, let's focus on the more challenging part: finding the derivative of ${\sin(x)^{\cos(x)}}$. We'll denote this part as ${u = \sin(x)^{\cos(x)}}$. The key idea behind logarithmic differentiation is to take the natural logarithm of both sides of the equation. This transforms the exponential expression into a more manageable form using logarithmic properties. Taking the natural logarithm of both sides of ${u = \sin(x)^{\cos(x)}}$, we get:

${ \ln(u) = \ln(\sin(x)^{\cos(x)}) }$ Using the property of logarithms that ${\ln(a^b) = b \ln(a)}$, we can rewrite the equation as:

${ \ln(u) = \cos(x) \ln(\sin(x)) }$ Now, we differentiate both sides with respect to ${x}$. On the left side, we apply the chain rule: ${ \frac{d}{dx}(\ln(u)) = \frac{1}{u} \frac{du}{dx} }$ On the right side, we have a product of two functions, ${\cos(x)}$ and ${\ln(\sin(x))}$, so we apply the product rule:

${ \frac{d}{dx}(\cos(x) \ln(\sin(x))) = \cos(x) \frac{d}{dx}(\ln(\sin(x))) + \ln(\sin(x)) \frac{d}{dx}(\cos(x)) }$ To differentiate ${\ln(\sin(x))}$, we use the chain rule again:

${ \frac{d}{dx}(\ln(\sin(x))) = \frac{1}{\sin(x)} \cdot \cos(x) = \frac{\cos(x)}{\sin(x)} = \cot(x) }$ The derivative of ${\cos(x)}$ is ${-\sin(x)}$. Substituting these results back into the equation, we get:

${ \frac{1}{u} \frac{du}{dx} = \cos(x) \cot(x) - \ln(\sin(x)) \sin(x) }$ Now, we solve for ${\frac{du}{dx}}$ by multiplying both sides by ${u}$:

${ \frac{du}{dx} = u(\cos(x) \cot(x) - \sin(x) \ln(\sin(x))) }$ Substituting ${u = \sin(x)^{\cos(x)}}$ back in, we get:

${ \frac{du}{dx} = \sin(x)^{\cos(x)} (\cos(x) \cot(x) - \sin(x) \ln(\sin(x))) }$ Finally, we add the derivative of ${x^2}$ to this result to find the derivative of the entire function ${y}$:

${ \frac{dy}{dx} = 2x + \sin(x)^{\cos(x)} (\cos(x) \cot(x) - \sin(x) \ln(\sin(x))) }$ In summary, this problem required a combination of the power rule for the ${x^2}$ term and logarithmic differentiation for the ${\sin(x)^{\cos(x)}}$ term. Logarithmic differentiation involved taking the natural logarithm of both sides, differentiating implicitly, and then solving for the derivative. This technique is crucial for handling functions where both the base and the exponent are functions of ${x}$.

Solution:

Let ${ y = x^2 + \sin(x)^{\cos(x)} }$

Let ${ u = \sin(x)^{\cos(x)} }$

Taking the natural logarithm:

${ \ln(u) = \cos(x) \ln(\sin(x)) }$ Differentiating implicitly:

${ \frac{1}{u} \frac{du}{dx} = \cos(x) \cdot \frac{\cos(x)}{\sin(x)} + \ln(\sin(x))(-\sin(x)) }$ ${ \frac{du}{dx} = u \left(\cos(x) \cot(x) - \sin(x) \ln(\sin(x))\right) }$ ${ \frac{du}{dx} = \sin(x)^{\cos(x)} \left(\cos(x) \cot(x) - \sin(x) \ln(\sin(x))\right) }$

${ \frac{dy}{dx} = 2x + \sin(x)^{\cos(x)} \left(\cos(x) \cot(x) - \sin(x) \ln(\sin(x))\right) }$