
How do you know which rule of differentiation to use?
Answer
535.2k+ views
Hint: Differentiation is the process of determining the derivation of a given function. There are different rules of differentiation used to solve the derivation. But here, I explain a few rules with an example: derivation is mentioned in the following.
Complete step by step solution:
General rule for differentiation,
\[\dfrac{d}{{dx}}[{x^n}] = n{x^{n - 1}}\], Where \[n \in R\;,n \ne 0.\]
The derivative of constant is zero,
\[\dfrac{d}{{dx}}[k] = 0\]
The derivative of a constant multiplied by a function is equal to the constant multiplied by the derivative of the function,
\[\dfrac{d}{{dx}}[k \cdot f(x)] = k\dfrac{d}{{dx}}[f(x)]\]
The derivative of a sum is equal to the sum of the derivatives,
\[\dfrac{d}{{dx}}[f(x) + g(x)] = \dfrac{d}{{dx}}[f(x)] + \dfrac{d}{{dx}}[g(x)]\]
The derivative of a difference is equal to the difference of the derivatives,
\[\dfrac{d}{{dx}}[f(x) - g(x)] = \dfrac{d}{{dx}}[f(x)] - \dfrac{d}{{dx}}[g(x)]\]
Some important rules are given the following,
Simple rule, \[f(x) = xdx = 1\]
Product rule, \[f(x) = - 4xdx = - 4\]
Power rule, \[f(x) = {x^2}dx = 2x\]
Chain rule, \[\dfrac{d}{{dx}}[f(g(x))] = f\prime (g(x))g\prime (x)\]
Example: Find the derivation of the function\[y = 2\sqrt x - 3\sin x\].
By using basic differentiation rule,
$
y\prime (x) = (2\sqrt x - 3sinx)\prime \\
\Rightarrow(2\sqrt x )\prime - (3sinx)\prime \\
\Rightarrow 2(\sqrt x )\prime - 3(sinx)\prime \\
\Rightarrow 2 \cdot \dfrac{1}{{2\sqrt x }} - 3cosx \\
= \dfrac{1}{{\sqrt x }} - 3cosx. \\
$
Hence, the final answer is found.
Note: One useful thing to keep in mind is that the derivative of a sum is the sum of the derivatives, so if you have more terms you can differentiate them one by one. The things you'll meet more often are powers of a function and most of all composed of functions. An example of the first may be\[\cos 2(x)\], an example of the second \[\log (2x)\] Of course both things could happen at the same time \[sin2{\text{ }}(4x + 2)\].However, differentiation in general way, you differentiate the outer content, and then the inner one, applying the basic rules for fundamental functions.
Complete step by step solution:
General rule for differentiation,
\[\dfrac{d}{{dx}}[{x^n}] = n{x^{n - 1}}\], Where \[n \in R\;,n \ne 0.\]
The derivative of constant is zero,
\[\dfrac{d}{{dx}}[k] = 0\]
The derivative of a constant multiplied by a function is equal to the constant multiplied by the derivative of the function,
\[\dfrac{d}{{dx}}[k \cdot f(x)] = k\dfrac{d}{{dx}}[f(x)]\]
The derivative of a sum is equal to the sum of the derivatives,
\[\dfrac{d}{{dx}}[f(x) + g(x)] = \dfrac{d}{{dx}}[f(x)] + \dfrac{d}{{dx}}[g(x)]\]
The derivative of a difference is equal to the difference of the derivatives,
\[\dfrac{d}{{dx}}[f(x) - g(x)] = \dfrac{d}{{dx}}[f(x)] - \dfrac{d}{{dx}}[g(x)]\]
Some important rules are given the following,
Simple rule, \[f(x) = xdx = 1\]
Product rule, \[f(x) = - 4xdx = - 4\]
Power rule, \[f(x) = {x^2}dx = 2x\]
Chain rule, \[\dfrac{d}{{dx}}[f(g(x))] = f\prime (g(x))g\prime (x)\]
Example: Find the derivation of the function\[y = 2\sqrt x - 3\sin x\].
By using basic differentiation rule,
$
y\prime (x) = (2\sqrt x - 3sinx)\prime \\
\Rightarrow(2\sqrt x )\prime - (3sinx)\prime \\
\Rightarrow 2(\sqrt x )\prime - 3(sinx)\prime \\
\Rightarrow 2 \cdot \dfrac{1}{{2\sqrt x }} - 3cosx \\
= \dfrac{1}{{\sqrt x }} - 3cosx. \\
$
Hence, the final answer is found.
Note: One useful thing to keep in mind is that the derivative of a sum is the sum of the derivatives, so if you have more terms you can differentiate them one by one. The things you'll meet more often are powers of a function and most of all composed of functions. An example of the first may be\[\cos 2(x)\], an example of the second \[\log (2x)\] Of course both things could happen at the same time \[sin2{\text{ }}(4x + 2)\].However, differentiation in general way, you differentiate the outer content, and then the inner one, applying the basic rules for fundamental functions.
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