How To Use Differentiation in Mathematics
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Differentiation is steps, in maths, We find the instantaneous rate of change in function based on one of its variables. The process finding derivatives is called differentiation.
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Differentiation is also called Derivative.
`\Rightarrow` `\frac{dy}{dx}` is read as dy by dx.
`\Rightarrow`Here `\frac{d}{dx}`(read d by dx) is an operator.
`\Rightarrow` `\frac{dy}{dx}` is not a quotient of by and dx.
`\Rightarrow` At a later stage we shall define dy as differential of y and dx as differential of x and hence `\frac{dy}{dx}` is also called differential coefficient.
Differentiation is steps, in maths, We find the instantaneous rate of change in function based on one of its variables. The process finding derivatives is called differentiation.
`\star`How to read differentiation:
`\Rightarrow` `\frac{dy}{dx}` is read as dy by dx.
`\Rightarrow`Here `\frac{d}{dx}`(read d by dx) is an operator.
`\Rightarrow` `\frac{dy}{dx}` is not a quotient of by and dx.
`\Rightarrow` At a later stage we shall define dy as differential of y and dx as differential of x and hence `\frac{dy}{dx}` is also called differential coefficient.
`\star` Formula:
`\Rightarrow``\frac{dy}{dx}` (k) = 0
`\Rightarrow``\frac{dy}{dx}` (x) = 1
`\Rightarrow``\frac{dy}{dx}` (`x^n`) = `n(x)^{n-1}`
`\Rightarrow``\frac{dy}{dx}` (log x) = `\frac{1}{x}`
`\Rightarrow``\frac{dy}{dx}` (`\frac{1}{x}`) = `\frac{-1}{x^2}`
`\Rightarrow``\frac{dy}{dx}` (`e^x`) = ex
`\Rightarrow``\frac{dy}{dx}` (`a^x`) = `a^x log_e a`
`\Rightarrow``\frac{dy}{dx}` (`\sqrt{x}`) = `\frac{1}{2\sqrt{x}}`
`\Rightarrow``\frac{dy}{dx}` (sin x) = cosx
`\Rightarrow``\frac{dy}{dx}` (cos x) = -sinx
`\Rightarrow``\frac{dy}{dx}` (tan x) = `sec^2 x`
`\Rightarrow``\frac{dy}{dx}` (cot x) = `-cosec^2 x`
`\Rightarrow``\frac{dy}{dx}` (sec x) = secx • tanx
`\Rightarrow``\frac{dy}{dx}` (cosec x) = -cosecx • cotx
`\Rightarrow``\frac{dy}{dx}` (`sin^{-1}x`) = `\frac{1}{\sqrt{1-x^{2}}}`
`\Rightarrow``\frac{dy}{dx}` (`cos^{-1}x`) = `\frac{-1}{\sqrt{1-x^{2}}}`
`\Rightarrow``\frac{dy}{dx}` (`tan^{-1}x`) = `\frac{1}{1+x^{2}}`
`\Rightarrow``\frac{dy}{dx}` (`cot^{-1}x`) = `\frac{-1}{1+x^{2}}`
`\Rightarrow``\frac{dy}{dx}` (`sec^{-1}x`) = `\frac{1}{|x|\sqrt{x^{2}-1}}`
`\Rightarrow``\frac{dy}{dx}` (`cosec^{-1}x`) = `\frac{-1}{|x|\sqrt{x^{2}-1}}`
`\star` Basic of Differentiation:
Some differentiation rules are a snap to remember and use. These include the Addition, Subtraction, Multiplication, Division, and Chain rule.
`\star`Addition and Substarction rule:
`\Rightarrow` Formula:
`\Rightarrow``\frac{d}{dx}(u+v) = \frac{d}{dx}u + \frac{d}{dx}v`
`\Rightarrow` Ex1: `Y_{1}` = log x + sin x
`\Rightarrow` Explanation:
step1:
`\Rightarrow`In this question we have find out derivative.
`\Rightarrow`According to question, we have to follow addtion and substarction rule.
`\Rightarrow`here, `\frac{d}{dx}(u+v) = \frac{d}{dx}u + \frac{d}{dx}v`
`\Rightarrow` We follow this rules. So, According to the rules, d/dx(u+v) Here, we have to contribute the question.
`\Rightarrow``\frac{d}{dx} y = \frac{d}{dx}(logx + sinx)`
step2:
In step 2 we have to slove step 1 so, rules is `\frac{d}{dx}u + \frac{d}{dx}v`
`\Rightarrow``\frac{dy}{dx} = \frac{d}{dx}logx + frac{d}{dx}sinx`
step3:
In step 3 `\frac{d}{dx}` logx = `\frac{1}{x}` and `\frac{d}{dx}` sinx = cosx
`\Rightarrow`So, our answer is that `\frac{1}{x}` + cosx
`\star`Extra sum in Addition and Substarction rule:
01) y= secx - x
02) y= `e^e + x^e + e^x`
03) `y= x^{\frac{4}{3}}`
04) y= (x-1)^2
`\star` Multiplication rule:
`\Rightarrow` Formula:
`\Rightarrow``\frac{d}{dx}(u•v) = u•\frac{d}{dx}•v + v•\frac{d}{dx}•u`
`\Rightarrow` Ex1: Y = logx • sin x
`\Rightarrow` Explanation:
step1:
`\Rightarrow`First we have to consider left hand side and right hand side.
step2:
`\Rightarrow`Left hand side and right hand side first we have to derivate left hand side(Left hand side is a y). Then derivation of y.
`\frac{dy}{dx} = \frac{d}{dx} (tanx • log)`
step3:
`\Rightarrow`Put value of y is step 1.
step4:
`\Rightarrow`Then meaning of step 3 is derivate of question with respect to x.
`\Rightarrow`We have to see in question use multiplication rule to slove it. Two method is available to for this sum.
First method:
`\Rightarrow` First term tanx is consider as u, Second term log is consider as v. (Then put the value multiplication rule which we have seen.)
Second method:
`\Rightarrow` First term is constant into derivation of second term. Then your first term tanx and second term is logx.
`\frac{dy}{dx} = tanx \frac{d}{dx} logx + logx \frac{d}{dx} tanx`
step2:
W.K.T.[We know that d/dx of logx means derivation of logx, first term as it's and after + sign , similary second logx is constant and means derivation of tanx.]
step3:
`\Rightarrow` tanx • `\frac{d}{dx}`• logx that means then we have to derivative holl term.
Answer: tanx • `\frac{1}{x}` + `sec^2x • tan^2x`
`\star`Extra sum in Multiplication rule:
01) y = `x^3` • sinx
02) y = `x^5` • `6^x`
03) y = x • logx
04) y = cotx • logx
`\star` Division rule:
`\Rightarrow` Formula:
`\Rightarrow``\frac{d}{dx} \frac{u}{v} = \frac{ v • \frac{d}{dx}• u - u • \frac{d}{dx}• v}{v^{2}}`
`\Rightarrow` Ex1: Y = `\frac{logx}{x}`
`\Rightarrow` Explanation:
`\Rightarrow`We have to see in question use Division rule to slove it. Two method is available to for this sum.
First method:
`\Rightarrow` First term logx is consider as u, Second term x is consider as v. (Then put the value multiplication rule which we have seen.)
Second method:
`\Rightarrow` First term is constant into derivation of second term. Then your first term x and second term is x.
`\frac{dy}{dx} = \frac{ x • \frac{d}{dx}• logx - logx • \frac{d}{dx}• x}{x^{2}} `
step2:
W.K.T.[We know that d/dx of logx means derivation of logx, first term as it's and after + sign , similary second logx is constant and means derivation of tanx. ]
= `\frac{x • \frac{1}{x} - log x (1)}{x^{2}}`
step3:
`\Rightarrow` x•`\frac{1}{x}` that means x , logx(1) that means logx.
Answer: `\frac{1-logx}{x^{2}}`
`\star`Extra sum in Division rule:
01) y = `\frac{4x +5}{2x -7}`
`\Rightarrow` Answer = `\frac{-638}{(2x -7)^{2}}`
02) y = `\frac{x^{2}-1}{x^{2} +1}`
`\Rightarrow` Answer = `\frac{4x}{(x^{2} +1)^{2}}`
03) y = `\frac{1+sinx}{1-sinx}`
`\Rightarrow` Answer = `\frac{2cosx}{(1 -sinx)^{2}}`
`\star` Chain rule:
`\Rightarrow` Explanation:
`\Rightarrow` When we use chain rule, When one or more function is in your question. Then use in chain rule.
`\Rightarrow` We have seen detail in ex1:
`\Rightarrow` Ex1: Y = `(3x -4)^{4}`
step1:
`\Rightarrow` Your first function is `x^n` x is consider is (3x -2) and second function is 4.
`\Rightarrow` Second function is x equation (3x-2) there is chain rule condition is satisfied in your question, That means one or more fuction is in youe question then we have use in chain rule.
= ` 4(3x-2)^{3}`•`\frac{d}{dx}`• (3x-2)
step2:
`\Rightarrow` We have to derivation of first function after second function.
=` 4(3x-2)^{3}` • (3-0)
= ` 4(3x-2)^{3}` • 3
= 12`(3x-2)^{3}`
`\star`Extra sum in Chain rule:
01) y = `\sqrt{sinx}`
`\Rightarrow` Answer = `\frac{1}{2\sqrt{sinx}}•cosx`
02) y = log(sinx)
`\Rightarrow` Answer = cotx
03) y = `e^{sinx + cosx}`
`\Rightarrow` Answer =`e^{sinx + cosx}`(cosx - sinx)
04) y = log(secx + tanx)
`\Rightarrow` Answer = secx
05) y = log `\frac{x^{2}+1}{x^{2}-1}`
`\Rightarrow` Answer = `\frac{-4x}{x^{4}-1}`
`star` Implict function:
`\Rightarrow` Explanation:
`\Rightarrow` In implicit differentiation, we differentiate each side of an equation with two variables (usually xxx and yyy) by treating one of the variables as a function of the other.
`\Rightarrow` Ex1: `x^{2} + xy + y^{2}` = 0
step1:
`\Rightarrow` Here, we treat y as an implicit function of x.
= `x^{2} + xy + y^{2}` = 0
= `\frac{d}{dx}`(`x^{2} + xy + y^{2}`) = 0
`\Rightarrow` In step 1 we have to slove Addition rule, rules is `\frac{d}{dx}u + \frac{d}{dx}v`
= `\frac{d}{dx}``x^{2}` + `\frac{d}{dx}` xy + `\frac{d}{dx}``y^{2}` + 2y `\frac{dy}{dx}`
step2:
`\Rightarrow` The first is term is d/dx into x res to 2 that means 2x and second term is d/dx into xy that means x into d/dx y and third term is y into d/dx into x that means y and four term is 2y into dy/dx that means as it's.
`\Rightarrow` 2x + x `\frac{dy}{dx}` + y + 2y `\frac{dy}{dx}` = 0
step3:
`\Rightarrow` dy/dx term is one side equal to simple term is another side.
= x `\frac{dy}{dx}` + 2y `\frac{dy}{dx}` = -2x -y
= (x + 2y) `\frac{dy}{dx}` = -2x +y
`\frac{dy}{dx}` = `frac{-2x+y}{x+2y}`
`\star`Extra sum in Implict function:
01) y = `x^{3}` + `y^{3}` = 3axy
`\Rightarrow` Answer = `\frac{ay-x^{2}}{y^{2}- ax}`
02) y = x • siny + y• sinx = 5
`\Rightarrow` Answer = `\frac{dy}{dx}` = `\frac{ycosx + sin}{xcosy + sinx}`
03) y = sinx (x+y)
`\Rightarrow` Answer = `\frac{cos x+ y}{1 - cos(x+y)}`
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