## How to calculation the derivative of sin^2x

*Note that in this write-up we will certainly be feather at differentiating sin2(x) i beg your pardon is not the same as separating sin(2x). Here is our write-up dealing with exactly how to distinguish sin(2x).You are watching: Derivative of sin(x/2)*

There are two approaches that have the right to be offered for calculating the derivative that sin^2x.

The very first method is by utilizing the product ascendancy for derivatives (since sin2(x) have the right to be composed as sin(x).sin(x)).

The second an approach is by utilizing the chain preeminence for differentiation.

### Finding the derivative the sin^2x making use of the product rule

The product preeminence for differentiation says that the derivative the f(x).g(x) is f’(x)g(x) + f(x).g’(x)

**The Product Rule:****For two differentiable attributes f(x) and also g(x)**

**If F(x) = f(x).g(x)**

**Then the derivative of F(x) is F"(x) = f’(x)g(x) + f(x)g"(x)**

First, let F(x) = sin2(x)

Then remember the sin2(x) is equal to sin(x).sin(x)

So F(x) = sin(x)sin(x)

By setting f(x) and also g(x) together sin(x) way that F(x) = f(x).g(x) and also we can use the product preeminence to find F"(x)

F"(x) | = f"(x)g(x) + f(x)g"(x) | Product preeminence Definition |

= f"(x)sin(x) + sin(x)g"(x) | f(x) = g(x) = sin(x) | |

= cos(x)sin(x) + sin(x)cos(x) | f"(x) = g(‘x) = cos(x) | |

= 2sin(x)cos(x) |

Using the product rule,**the derivative of sin^2x is 2sin(x)**cos(x)

### Finding the derivative the sin^2x making use of the chain rule

The chain preeminence is valuable for recognize the derivative that a duty which can have been distinguished had it remained in x, but it is in the form of another expression which could additionally be differentiated if that stood ~ above its own.

In this case:

We know just how to distinguish sin(x) (the price is cos(x))We know exactly how to identify x2 (the answer is 2x)This method the chain ascendancy will allow us to do the differentiation that the expression sin^2x.

Using the chain dominion to discover the derivative the sin^2xAlthough the expression sin2x contains no parenthesis, we can still see it as a composite role (a function of a function).

We have the right to write sin2x together (sin(x))2.

Now the role is in the form of x2, except it go not have x as the base, rather it has actually another role of x (sin(x)) as the base.

Let’s call the duty of the base g(x), i m sorry means:

g(x) = sin(x)

From this it follows that:

sin(x)2 = g(x)2

So if the role f(x) = x2 and the duty g(x) = sin(x), then the duty (sin(x))2 can be composed as a composite function.

f(x) = x2f(g(x)) = g(x)2 (but g(x) = sin(x))f(g(x)) = (sin(x))2

Let’s specify this composite function as F(x):

F(x) = f(g(x)) = (sin(x))2

We can uncover the derivative of sin^2x (F"(x)) by making usage of the chain rule.

**The Chain Rule:****For 2 differentiable features f(x) and g(x)**

**If F(x) = f(g(x))**

**Then the derivative the F(x) is F"(x) = f’(g(x)).g’(x)**

Now we deserve to just plug f(x) and g(x) into the chain rule.

**How to find the derivative the sin^2x using the Chain Rule:**

F"(x) | = f"(g(x)).g"(x) | Chain dominion Definition |

= f"(g(x))(cos(x)) | g(x) = sin(x) ⇒ g"(x) = cos(x) | |

= (2sin(x)).(cos(x)) | f(g(x)) = (sin(x))2 ⇒ f"(g(x)) = 2sin(x) | |

= 2sin(x)cos(x) |

Using the chain rule,**the derivative the sin^2x is 2sin(x)cos(x)**

*(Note – making use of the trigonometric identity 2cos(x)sin(x) = sin(2x), the derivative that sin^2x can also be created as sin(2x))*

Finally, simply a note on syntax and also notation: sin^2x is sometimes written in the forms listed below (with the derivative together per the calculations above). Just be conscious that not every one of the forms listed below are mathematically correct.

sin2x | ► Derivative the sin2x = 2sin(x)cos(x) |

sin^2(x) | ► Derivative of sin^2(x) = 2sin(x)cos(x) |

sin 2 x | ► Derivative the sin 2 x = 2sin(x)cos(x) |

(sinx)^2 | ► Derivative of (sinx)^2 = 2sin(x)cos(x) |

sin squared x | ► Derivative of sin squared x = 2sin(x)cos(x) |

sinx2 | ► Derivative that sinx2 = 2sin(x)cos(x) |

sin^2 | ► Derivative the sin^2 = 2sin(x)cos(x) |

## The second Derivative that sin^2x

To calculate the 2nd derivative the a function, girlfriend just identify the very first derivative.

From above, we discovered that the an initial derivative the sin^2x = 2sin(x)cos(x). So to discover the second derivative that sin^2x, we just need to identify 2sin(x)cos(x)

We have the right to use the product rule to discover the derivative of 2sin(x)cos(x).

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We can collection f(x) = 2sin(x) and g(x) = cos(x) and also apply the product dominion to find the derivative that f(x).g(x) = 2sin(x)cos(x)

The product rules claims that the derivative that f(x).g(x) is equal to f’(x)g(x) + f(x)g’(x).

2cos(x)cos(x) + 2sin(x)(-sin(x))= 2cos2(x) – 2sin2(x)= 2(cos2(x) – sin2(x))

Using the trigonometric double angle identity cos(2x) = cos2(x) – sin2(x), we have the right to rewrite this as

= 2cos(2x)

► **The second derivative the sin^2x is 2cos(2x)**

*Interestingly, the second derivative of sin2x is equal to the first derivative that sin(2x). *