第11类RD Sharma解决方案-第30章导数–练习30.1

问题1.在x = 2处找到f(x)= 3x的导数

解决方案：

Given: f(x)=3x

By using the derivative formula,

$f'(a)= \lim_{h \to 0} \frac {f(a+h)-f(a)} h$ {where h is a small positive number}

Derivative of f(x)=3x at x=2 is given as:

$f'(2)= \lim_{h \to 0} \frac {f(2+h)-f(2)} h$

⇒ $f'(2)= \lim_{h \to 0} \frac {3(2+h)-3(2)} h$

⇒ $f'(2)= \lim_{h \to 0} \frac {3h+6-6} h$

⇒ $f'(2)= \lim_{h \to 0} \frac {3h} h$

⇒ $f'(2)= \lim_{h \to 0} 3 = 3$

Hence, derivative of f(x)=3x at x=2 is 3

问题2.在x = 10时找到f(x)= x ^{2 – 2的导数}

解决方案：

Given: f(x)= x²-2 

By using the derivative formula,

$f'(a)= \lim_{h \to 0} \frac {f(a+h)-f(a)} h$ {where h is a small positive number}

Derivative of f(x)=x²-2 at x=10 is given as:

$f'(10)= \lim_{h \to 0} \frac {f(10+h)-f(10)} h$

⇒$f'(10)= \lim_{h \to 0} \frac {(100+h^2+20h-2-100+2))} h$

⇒$f'(10)= \lim_{h \to 0} \frac {h^2+20h} h$

⇒$f'(10)= \lim_{h \to 0} \frac {h(h+20)} h$

⇒$f'(10)= \lim_{h \to 0} {h+20}$

⇒$f'(10)= 0+20=20$

Hence, derivative of f(x)=x²-2 at x=10 is 20

问题3.在x = 100时找到f(x)= 99x的导数

解决方案：

Given: f(x)= 99x

By using the derivative formula,

$f'(a)= \lim_{h \to 0} \frac {f(a+h)-f(a)} h$ {where h is a small positive number}

Derivative of f(x)=99x at x=100 is given as:

$f'(100)= \lim_{h \to 0} \frac {f(100+h)-f(100)} h$

⇒ $f'(100)= \lim_{h \to 0} \frac {99(100+h)-99(100)} h$

⇒ $f'(100)= \lim_{h \to 0} \frac {9900+99h-9900} h$

⇒ $f'(100)= \lim_{h \to 0} 99 = 99$

Hence, derivative of f(x)=99x at x=100 is 99

问题4.在x = 1处找到f(x)= x的导数

解决方案：

Given: f(x)=x

By using the derivative formula,

$f'(a)= \lim_{h \to 0} \frac {f(a+h)-f(a)} h${where h is a small positive number}

Derivative of f(x)=x at x=1 is given as:

$f'(1)= \lim_{h \to 0} \frac {f(1+h)-f(1)} h$

⇒ $f'(1)= \lim_{h \to 0} \frac {(1+h)-1} h$

⇒ $f'(1)= \lim_{h \to 0} \frac {h} h$

⇒$f'(1)= \lim_{h \to 0} 1 = 1$

Hence, derivative of f(x)=x at x=1 is 1

问题5.求f(x)=的导数$\cos x$在x = 0

解决方案：

Given: f(x)=$\cos x$

By using the derivative formula,

$f'(a)= \lim_{h \to 0} \frac {f(a+h)-f(a)} h${where h is a small positive number}

Derivative of f(x)=$\cos x$ at x=0 is given as:

$f'(0)= \lim_{h \to 0} \frac {f(0+h)-f(0)} h$

⇒ $f'(0)= \lim_{h \to 0} \frac {\cos(0+h)-\cos(0)} h$

⇒ $f'(0)= \lim_{h \to 0} \frac {\cos(h)-\cos(0)} h$

⇒ $f'(0)= \lim_{h \to 0} \frac {\cos(h)-1} h$

∵ we can not find the limit of the above function f(x)=$\cos x$ by direct substitution as it gives 0/0 form (indeterminate form)

So we will simplify it to find the limit.

As we know that $1 - \cos x = 2 \sin2(x/2)$

∴ $f'(0)= \lim_{h \to 0} \frac {-(1-\cos(h))} h = - \lim_{h \to 0} \frac {2\sin^2(h/2)} h$

Divide the numerator and denominator by 2 to get the form $\sin x/x$ for applying sandwich theorem and multiplying h in numerator and denominator to get the required form.

⇒ $f'(0)= -\lim_{h \to 0}( \frac {\frac {2\sin^2(h/2)}2} {\frac {h^2}2}) \times h$

⇒ $f'(0)= -\lim_{h \to 0}( \frac {\sin(h/2)} {\frac {h}2}) \times \lim_{h \to 0}h$

Using the formula: $\lim_{x \to 0} \frac{\sin x}x=1$

∴ $f'(0)=-1 \times 0=0$

Hence, derivative of f(x)=$\cos x$ at x=0 is 0

问题6.求f(x)=的导数$\tan x$在x = 0

解决方案：

Given: f(x)=$\tan x$

By using the derivative formula,

$f'(a)= \lim_{h \to 0} \frac {f(a+h)-f(a)} h${where h is a small positive number}

Derivative of f(x)=$\tan x$ at x=0 is given as:

$f'(0)= \lim_{h \to 0} \frac {f(0+h)-f(0)} h$

⇒ $f'(0)= \lim_{h \to 0} \frac {\tan(0+h)-\tan(0)} h$

⇒ $f'(0)= \lim_{h \to 0} \frac {\tan(h)-0} h$

⇒ $f'(0)= \lim_{h \to 0} \frac {\tan(h)} h$

∴ Use the formula: $\lim_{x \to 0} \frac {\tan x} x=1$ {sandwich theorem}

⇒ $f'(0)=1$

Hence, derivative of f(x)=$\tan x$ at x=0 is 1

问题7(i)。在指示的点找到以下函数的导数： $\sin x$在$x=\pi/2$

解决方案：

Given: f(x)= $\sin x$

By using the derivative formula,

$f'(a)= \lim_{h \to 0} \frac {f(a+h)-f(a)} h${where h is a small positive number}

Derivative of f(x)=$\sin x$ at $x=\pi/2$ is given as:

$f'(\pi/2)= \lim_{h \to 0} \frac {f(\pi/2+h)-f(\pi/2)} h$

⇒$f'(\pi/2)= \lim_{h \to 0} \frac {\sin(\pi/2+h)-\sin(\pi/2)} h$

⇒ $f'(\pi/2)= \lim_{h \to 0} \frac {\sin(\pi/2+h)-1} h$

⇒  f'(\pi/2)= $\lim_{h \to 0} \frac {\cos(h)-1} h$  {∵$\sin(\pi/2+x)=\cos x$

∵ we can not find the limit of the above function by direct substitution as it gives 0/0 form (indeterminate form)

So we will simplify it to find the limit.

As we know that

$1 - \cos x = 2 \sin^2(x/2)$

∴ $f'(\pi/2)= \lim_{h \to 0} \frac {-(1-\cos(h))} h = - \lim_{h \to 0} \frac {2\sin^2(h/2)} h$

Divide the numerator and denominator by 2 to get the form (sin x)/x for applying sandwich theorem and multiplying h in numerator and denominator to get the required form.

⇒ $f'(\pi/2)= -\lim_{h \to 0}( \frac {\frac {2\sin^2(h/2)}2} {\frac {h^2}2}) \times h$

⇒ $f'(\pi/2)= -\lim_{h \to 0}( \frac {\sin(h/2)} {\frac {h}2}) \times \lim_{h \to 0}h$

Using the formula:

$\lim_{x \to 0} \frac{\sin x}x=1$

∴$f'(\pi/2)=-1 \times 0=0$

Hence, derivative of f(x)= $\sin x$ at $x=\pi/2$ is 0

问题7(ii)。在指示的点找到以下函数的导数：x在x = 1处

解决方案：

Given: f(x)=x

By using the derivative formula,

$f'(a)= \lim_{h \to 0} \frac {f(a+h)-f(a)} h$ {where h is a small positive number}

Derivative of f(x)=x at x=1 is given as:

$f'(1)= \lim_{h \to 0} \frac {f(1+h)-f(1)} h$

⇒ $f'(1)= \lim_{h \to 0} \frac {(1+h)-1} h$

⇒ $f'(1)= \lim_{h \to 0} \frac {h} h$

⇒ $f'(1)= \lim_{h \to 0} 1 = 1$

Hence, derivative of f(x)=x at x=1 is 1

问题7(iii)。在指示的点找到以下函数的导数： 2 \ cos x at $x=\pi/2$

解决方案：

Given: f(x)= $2\cos x$

By using the derivative formula,

$f'(a)= \lim_{h \to 0} \frac {f(a+h)-f(a)} h$ {where h is a small positive number}

Derivative of f(x)= $2\cos x$ at $x=\pi/2$ is given as:

$f'(\pi/2)= \lim_{h \to 0} \frac {f(\pi/2+h)-f(\pi/2)} h$

⇒ $f'(\pi/2)= \lim_{h \to 0} \frac {2\cos(\pi/2+h)-2cos(\pi/2)} h$

⇒ f'(\pi/2)= \lim_{h \to 0} \frac {-2\sin(h)} h {∵ $\cos(\pi/2+x)=-\sin x$}

∵ we can not find the limit of the above function by direct substitution as it gives 0/0 form (indeterminate form)

∴ $f'(\pi/2)= -2\lim_{h \to 0} \frac {\sin h} h$

Using the formula: $\lim_{x \to 0} \frac{\sin x}x=1$

∴ $f'(\pi/2)=-2 \times 1=-2$

Hence, derivative of f(x)= $2\cos x =-2$

问题7(iv)。在指示的点找到以下函数的导数： $\sin 2x$在$x=\pi/2$

解决方案：

Given: f(x)= $\sin 2x$

By using the derivative formula,

$f'(a)= \lim_{h \to 0} \frac {f(a+h)-f(a)} h$ {where h is a small positive number}

Derivative of f(x)= $\sin 2x$ at $x=\pi/2$ is given as:

$f'(\pi/2)= \lim_{h \to 0} \frac {f(\pi/2+h)-f(\pi/2)} h$

⇒ $f'(\pi/2)= \lim_{h \to 0} \frac {\sin 2(\pi/2+h)-sin 2(\pi/2)} h$

⇒ $f'(\pi/2)= \lim_{h \to 0} \frac {\sin(\pi+2h) - \sin \pi} h$ {∵$\sin(\pi+x)=-sinx$}

⇒ $f'(\pi/2)= \lim_{h \to 0} \frac {-\sin(2h) - 0} h$

⇒ $f'(\pi/2)= \lim_{h \to 0} \frac {-\sin(2h)} h$

∵ we can not find the limit of the above function by direct substitution as it gives 0/0 form (indeterminate form)

Using the sandwich theorem  and multiplying 2 in numerator and denominator to apply the formula.

$f'(\pi/2)= -\lim_{h \to 0} \frac {\sin(2h)} {2h} \times 2 = -2\lim_{h \to 0} \frac {\sin(2h)} {2h}$

Using the formula: $\lim_{x \to 0} \frac{\sin x}x=1$

∴ $f'(\pi/2)=-2 \times 1=-2$

Hence, derivative of f(x)=$\sin 2x=-2$