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คำตอบ - Derivative

(ec×ddx[c])×osx+ec×ddx[o]×sx+ec×o×ddx[s]×x+osec
(e^{c}\times \frac{d}{dx}[c])\times osx+e^{c}\times \frac{d}{dx}[o]\times sx+e^{c}\times o\times \frac{d}{dx}[s]\times x+o s e^{c}

Other Ways to Solve

Derivative

คำอธิบายทีละขั้นตอน

1. ขั้นตอน 160: จัดรูปอนุพันธ์

19 additional steps

Expanding the derivative for multiplication.

ddx[ec×osx]=ddx[ec]×osx+ec×ddx[o]×sx+ec×o×ddx[s]×x+ec×os×ddx[x]

Expanding the derivative for multiplication.

ddx[ec×osx]=ddx[ec]×osx+ec×ddx[o]×sx+ec×o×ddx[s]×x+ec×os×ddx[x]

Multiplication can be grouped differently, but the result remains the same.

ddx[ec×osx]=ddx[ec×(osx)]

Applying the product rule of derivatives.

ddx[ec×(osx)]=ddx[ec]×(osx)+ec×ddx[osx]

Expanding the derivative for multiplication.

ddx[ec×osx]=ddx[ec]×osx+ec×ddx[o]×sx+ec×o×ddx[s]×x+ec×os×ddx[x]

Expanding the derivative for multiplication.

ddx[ec]×(osx)+ec×ddx[osx]=ddx[ec]×(osx)+ec×(ddx[o]×sx+o×ddx[s]×x+os×ddx[x])

Multiplication can be grouped differently, but the result remains the same.

ddx[osx]=ddx[o×(sx)]

Applying the product rule of derivatives.

ddx[o×(sx)]=ddx[o]×(sx)+o×ddx[sx]

Expanding the derivative for multiplication.

ddx[ec]×(osx)+ec×ddx[osx]=ddx[ec]×(osx)+ec×(ddx[o]×sx+o×ddx[s]×x+os×ddx[x])

Applying the product rule of derivatives.

ddx[sx]=ddx[s]×x+s×ddx[x]

Multiplication can be grouped differently, but the result remains the same.

ddx[o]×(sx)+o(ddx[s]×x+s×ddx[x])=ddx[o]×sx+o(ddx[s]×x+s×ddx[x])

Multiplying a number by a sum or difference of two numbers can be done by multiplying each number individually and then adding or subtracting the results.

ddx[o]×sx+o(ddx[s]×x+s×ddx[x])=ddx[o]×sx+(o×(ddx[s]×x)+o×(s×ddx[x]))

Multiplication can be grouped differently, but the result remains the same.

ddx[o]×sx+(o×(ddx[s]×x)+o×(s×ddx[x]))=ddx[o]×sx+(o×ddx[s]×x+o×(s×ddx[x]))

Multiplication can be grouped differently, but the result remains the same.

ddx[o]×sx+(o×ddx[s]×x+o×(s×ddx[x]))=ddx[o]×sx+(o×ddx[s]×x+os×ddx[x])

Addition can be grouped differently, but the result remains the same.

ddx[o]×sx+(o×ddx[s]×x+os×ddx[x])=ddx[o]×sx+o×ddx[s]×x+os×ddx[x]

Multiplication can be grouped differently, but the result remains the same.

ddx[ec]×(osx)+ec×(ddx[o]×sx+o×ddx[s]×x+os×ddx[x])=ddx[ec]×osx+ec×(ddx[o]×sx+o×ddx[s]×x+os×ddx[x])

Multiplying a number by a sum or difference of two numbers can be done by multiplying each number individually and then adding or subtracting the results.

ddx[ec]×osx+ec×(ddx[o]×sx+o×ddx[s]×x+os×ddx[x])=ddx[ec]×osx+(ec×(ddx[o]×sx)+ec×(o×ddx[s]×x)+ec×(os×ddx[x]))

Multiplication can be grouped differently, but the result remains the same.

ddx[ec]×osx+(ec×(ddx[o]×sx)+ec×(o×ddx[s]×x)+ec×(os×ddx[x]))=ddx[ec]×osx+(ec×ddx[o]×sx+ec×(o×ddx[s]×x)+ec×(os×ddx[x]))

Multiplication can be grouped differently, but the result remains the same.

ddx[ec]×osx+(ec×ddx[o]×sx+ec×(o×ddx[s]×x)+ec×(os×ddx[x]))=ddx[ec]×osx+(ec×ddx[o]×sx+ec×o×ddx[s]×x+ec×(os×ddx[x]))

Multiplication can be grouped differently, but the result remains the same.

ddx[ec]×osx+(ec×ddx[o]×sx+ec×o×ddx[s]×x+ec×(os×ddx[x]))=ddx[ec]×osx+(ec×ddx[o]×sx+ec×o×ddx[s]×x+ec×os×ddx[x])

Addition can be grouped differently, but the result remains the same.

ddx[ec]×osx+(ec×ddx[o]×sx+ec×o×ddx[s]×x+ec×os×ddx[x])=ddx[ec]×osx+ec×ddx[o]×sx+ec×o×ddx[s]×x+ec×os×ddx[x]

Computing the derivative of a power function.

ddx[ec]×osx+ec×ddx[o]×sx+ec×o×ddx[s]×x+ec×os×ddx[x]=(ec×(ddx[c]×ln(e)+ce×ddx[e]))×osx+ec×ddx[o]×sx+ec×o×ddx[s]×x+ec×os×ddx[x]

The derivative of a variable with respect to itself is always equal to one.

(ec×(ddx[c]×ln(e)+ce×ddx[e]))×osx+ec×ddx[o]×sx+ec×o×ddx[s]×x+ec×os×ddx[x]=(ec×(ddx[c]×ln(e)+ce×ddx[e]))×osx+ec×ddx[o]×sx+ec×o×ddx[s]×x+ec×os×1

The derivative of a constant value is always zero.

(ec×(ddx[c]×ln(e)+ce×ddx[e]))×osx+ec×ddx[o]×sx+ec×o×ddx[s]×x+ec×os×1=(ec×(ddx[c]×ln(e)+ce×0))×osx+ec×ddx[o]×sx+ec×o×ddx[s]×x+ec×os×1

Simplifying the arithmetic expressions.

(ec×(ddx[c]×ln(e)+ce×0))×osx+ec×ddx[o]×sx+ec×o×ddx[s]×x+ec×os×1=(ec×(ddx[c]×1+ce×0))×osx+ec×ddx[o]×sx+ec×o×ddx[s]×x+ec×os×1

Multiplying a number by zero always results in zero.

(ec×(ddx[c]×1+ce×0))×osx+ec×ddx[o]×sx+ec×o×ddx[s]×x+ec×os×1=(ec×(ddx[c]×1+0))×osx+ec×ddx[o]×sx+ec×o×ddx[s]×x+ec×os×1

Simplifying the arithmetic expressions.

(ec×(ddx[c]×1+0))×osx+ec×ddx[o]×sx+ec×o×ddx[s]×x+ec×os×1=(ec×(ddx[c]×1+0))×osx+ec×ddx[o]×sx+ec×o×ddx[s]×x+osec

Adding zero to a number, which does not change its value.

(ec×(ddx[c]×1+0))×osx+ec×ddx[o]×sx+ec×o×ddx[s]×x+osec=(ec×(ddx[c]×1))×osx+ec×ddx[o]×sx+ec×o×ddx[s]×x+osec

Multiplying a number by one, which does not change its value.

(ec×(ddx[c]×1))×osx+ec×ddx[o]×sx+ec×o×ddx[s]×x+osec=(ec×ddx[c])×osx+ec×ddx[o]×sx+ec×o×ddx[s]×x+osec

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