The differentiation of (\\cos(x)) with respect to (x) is a fundamental concept in calculus. The derivative of (\\cos(x)) is given by:<\/p>\n\n\n\n
\\[
\\frac{d}{dx}(\\cos(x)) = -\\sin(x).
\\]<\/p>\n\n\n\n
This result is derived from the limit definition of the derivative and is a key part of the differentiation rules for trigonometric functions. The negative sine function, $-\\sin(x)$, represents the rate of change of $\\cos(x)$ with respect to (x).<\/p>\n\n\n\n
We can proof using the below two formulas<\/p>\n\n\n\n
$\\frac {d}{dx} f(x) =\\displaystyle \\lim_{h \\to 0} \\frac {f (x+h) – f(x)}{h}$<\/p>\n\n\n\n
$\\displaystyle \\lim_{x \\to 0} \\frac {\\sin(x)}{x}=1$<\/p>\n\n\n\n
Now
$\\frac {d}{dx} (\\cos x) = \\displaystyle \\lim_{h \\to 0} \\frac {\\cos (x+h) – \\cos x}{h}= \\displaystyle \\lim_{h \\to 0} -\\frac {2\\sin (\\frac {2x+h}{2}) sin \\frac {h}{2}}{h}$
$=-\\displaystyle \\lim_{h \\to 0} \\sin (x + \\frac {h}{2}) \\displaystyle \\lim_{h \\to 0} \\frac {sin \\frac {h}{2}}{\\frac {h}{2}} =- \\sin x$
as $\\displaystyle \\lim_{x \\to 0} \\frac {sinx}{x}=1$<\/p>\n\n\n\n
\\[
\\frac{d}{dx}(\\cos(x)) =- \\sin(x).
\\]<\/p>\n\n\n\n
We can clearly see that Domain of the Derivative of cos x is R<\/strong><\/p>\n\n\n\n The geometric meaning of the derivative of $\\cos(x)$, which is $-\\sin(x)$, can be understood by examining the graph of $\\cos(x)$ and considering what the derivative represents.<\/p>\n\n\n\n Question 1:<\/strong> Question 2:<\/strong> \\[ Question 3:<\/strong> \\[ Question 4<\/strong> Solution<\/strong> The slope of the tangent line at $x = \\frac{\\pi}{4}$ is $-\\frac{\\sqrt{2}}{2}$. The y-coordinate at this point is $\\cos(\\frac{\\pi}{4}) = \\frac{\\sqrt{2}}{2}$. Using the point-slope form of the equation of a line: Question 5<\/strong> Solution<\/strong> Solving for $x$, we get: Within the interval $[0, 2\\pi]$, the relevant values of $x$ are $0$, $\\pi$, and $2\\pi$. At $x = 0$ and $x = 2\\pi$, $\\cos(x) = 1$, indicating maximum values. At $x = \\pi$, $\\cos(x) = -1$, indicating a minimum value.<\/p>\n\n\n\n Question 6<\/strong> Solution<\/strong><\/p>\n\n\n\n Knowing that $\\cos(x) = \\sin(x + \\frac{\\pi}{2})$, we can differentiate both sides with respect to $x$: The derivative of the right side using the chain rule is: Since $\\cos(x + \\frac{\\pi}{2}) = \\cos(x)$’s derivative is $-\\sin(x)$, and by the periodicity and phase shift properties of sine and cosine, this demonstrates the relationship and the derivative of $\\sin(x)$ indirectly.<\/p>\n","protected":false},"excerpt":{"rendered":" The differentiation of (\\cos(x)) with respect to (x) is a fundamental concept in calculus. The derivative of (\\cos(x)) is given by: \\[\\frac{d}{dx}(\\cos(x)) = -\\sin(x).\\] This result is derived from the limit definition of the derivative and is a key part of the differentiation rules for trigonometric functions. The negative sine function, $-\\sin(x)$, represents the rate […]<\/p>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_uag_custom_page_level_css":"","site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"set","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[498],"tags":[],"class_list":["post-8972","post","type-post","status-publish","format-standard","hentry","category-maths"],"yoast_head":"\nGeometrical Meaning of the Derivative <\/h2>\n\n\n\n
\n
Solved Questions<\/h2>\n\n\n\n
Differentiate $y = \\cos(3x)$.
Solution:<\/strong>
To differentiate $y = \\cos(3x)$, we use the chain rule. Let (u = 3x), then $y = \\cos(u)$. Differentiating (y) with respect to (x):
\\[
\\frac{dy}{dx} = \\frac{dy}{du} \\cdot \\frac{du}{dx} = -\\sin(u) \\cdot 3 = -3\\sin(3x).
\\]<\/p>\n\n\n\n
Find the derivative of $y = x \\cos(x)$.
Solution<\/strong>
Here, we use the product rule. If (y = uv) where (u = x) and $v = \\cos(x)$, then (y’ = u’v + uv’). Therefore:<\/p>\n\n\n\n
\\frac{dy}{dx} = 1 \\cdot \\cos(x) – x \\cdot \\sin(x) = \\cos(x) – x\\sin(x).
\\]<\/p>\n\n\n\n
What is the second derivative of $y = \\cos(x)$?
Solution:<\/strong>
First, we find the first derivative, which is $y’ = -\\sin(x)$. Then, differentiating again:<\/p>\n\n\n\n
y” = \\frac{d}{dx}(\\cos(x)) = -\\cos(x).
\\]<\/p>\n\n\n\n
Find the equation of the tangent line to the curve $y = \\cos(x)$ at $x = \\frac{\\pi}{4}$.<\/p>\n\n\n\n
First, find the derivative of $y = \\cos(x)$, which is $y’ = -\\sin(x)$.
Then, evaluate this derivative at $x = \\frac{\\pi}{4}$ to find the slope of the tangent line:
$$y'(\\frac{\\pi}{4}) = -\\sin(\\frac{\\pi}{4}) = -\\frac{\\sqrt{2}}{2}$$<\/p>\n\n\n\n
$$y – \\frac{\\sqrt{2}}{2} = -\\frac{\\sqrt{2}}{2}(x – \\frac{\\pi}{4})$$<\/p>\n\n\n\n
Determine the $x$ values where the function $f(x) = \\cos(x)$ has maximum and minimum values on the interval $[0, 2\\pi]$.<\/p>\n\n\n\n
The derivative of $f(x) = \\cos(x)$ is $f'(x) = -\\sin(x)$. Setting the derivative equal to 0 to find critical points:
$$-\\sin(x) = 0$$<\/p>\n\n\n\n
$$x = n\\pi, \\text{ where } n \\text{ is an integer}$$<\/p>\n\n\n\n
Prove that the derivative of $\\sin(x)$ is $\\cos(x)$ using the derivative of $\\cos(x)$.<\/p>\n\n\n\n
$$f(x) = \\sin(x + \\frac{\\pi}{2})$$<\/p>\n\n\n\n
$$f'(x) = \\cos(x + \\frac{\\pi}{2}) \\cdot \\frac{d}{dx}(x + \\frac{\\pi}{2}) = \\cos(x + \\frac{\\pi}{2})$$<\/p>\n\n\n\n