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Chapter 6: Problem 7

Find the derivative of \(y=f(x)=\sin \left(x^{2}\right)\), using first principle.

Short Answer

Expert verified
The derivative of the function \(y=f(x)=\sin(x^{2})\) using the first principles is \(f'(x) = 2x\cos(2x^{2})\).

Step by step solution

01

Write down the function and the definition of the derivative

The given function is \(f(x) = \sin(x^{2})\). The limit definition of the derivative is \(f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\). Replace \(f(x+h)\) and \(f(x)\) with appropriate expressions, so the task now is to compute \(f'(x) = \lim_{h \to 0} \frac{\sin((x+h)^{2}) - \sin(x^{2})}{h}\).
02

Apply the sine difference identity

Use the trigonometric identity for the difference of two sines. This identity is \(\sin(A) - \sin(B) = 2 \cos \left( \frac{A + B}{2} \right) \sin \left( \frac{A - B}{2} \right)\). In this case, \(A = (x+h)^{2}\) and \(B = x^{2}\). After applying the identity, the expression becomes: \(f'(x) = \lim_{h \to 0} \frac{2 \cos \left( \frac{(x+h)^{2} + x^{2}}{2} \right) \sin \left( \frac{(x+h)^{2} - x^{2}}{2} \right)}{h}\).
03

Simplify the expression

It is noticed that \((x+h)^{2} + x^{2} = 2x^{2} + 2hx + h^{2}\) and \((x+h)^{2} - x^{2} = 2hx + h^{2}\). Replace these relations into our derivative expression, and separate the limit to apply the limit properties: \(f'(x) = \lim_{h \to 0} 2 \cos(2x^{2} + hx) \cdot \frac{\sin(hx +h^{2})}{h}\).
04

Use the small angle approximation

One of the limit properties is that the limit of the product is the product of the limits. In this case, the limit on the right is \(\sin(hx +h^{2}) / h\) which, considering that \(h \rightarrow 0\), may be approximated by \(x + h\), according to small angle approximation for the sine. This gives us: \(f'(x) = 2 \cos(2x^{2}) \cdot (x+h)\).
05

Compute the limit

Finally, letting \(h\) go to 0 gives us the derivative \(f'(x) = 2x \cos(2x^{2})\).

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Most popular questions from this chapter

If \(y=x^{n-1} \ln x\), then prove that $$ x^{2}\left(\frac{d^{2} y}{d x^{2}}\right)+(3-2 n) x \frac{d y}{d x}+(n-1)^{2} y=0 $$

If \(y=\tan ^{-1} x\), then prove that (i) \(\left(1+x^{2}\right) y_{2}+2 x y_{1}=0\) (ii) \(\left(1+x^{2}\right) y_{n+2}+2(n+2) x y_{n+1}+n(n+1) y_{n}\) \(=0 .\)

Find \(\frac{d^{2} y}{d x^{2}}\), if (i) \(x=a t^{2}, y=2 a t\) (ii) \(x=a \cos ^{3} \theta, y=a \sin ^{3} \theta\) (iii) \(x=a \cos \theta, y=b \sin \theta\)

If \(y=\cos ^{-1}\left(\sqrt{\frac{\cos 3 x}{\cos ^{3} x}}\right)\), prove that \(\frac{d y}{d x}=\sqrt{\frac{6}{\cos 2 x+\cos 4 x}}\)

If \(y=\left(C_{1}+C_{2} x\right) \sin x+\left(C_{3}+C_{4} x\right) \cos x\), show that \(\frac{d^{4} y}{d x^{4}}+2 \frac{d^{2} y}{d x^{2}}+y=0\)
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