Then the value of ${d \over {d\left( {\tan \theta } \right)}}\left( {f\left( \theta \right)} \right)$ is
Explanation:
$\sin \left( {{{\tan }^{ - 1}}\left( {{{\sin \theta } \over {\sqrt {\cos 2\theta } }}} \right)} \right)$, where $\theta \in \left( { - {\pi \over 4},{\pi \over 4}} \right)$
$\sin \left( {{{\tan }^{ - 1}}\left( {{{\sin \theta } \over {\sqrt {2{{\cos }^2}\theta - 1} }}} \right)} \right) = \sin ({\sin ^{ - 1}}(\tan \theta )) = \tan \theta $
${{d(\tan \theta )} \over {d(\tan \theta )}} = 1$
function such that $\left| {f\left( x \right)} \right| \le 1$ and $f'(x)=g(x).$
If ${f^2}\left( 0 \right) + {g^2}\left( 0 \right) = 9.$ Prove that there exists some $c \in \left( { - 3,3} \right)$
such that $g(c).g''(c)<0.$
If $f(x)$ is a differentiable function and $g(x)$ is a double differentiable function such that $|f(x)| \leq 1$ and $f'(x)=g(x)$, where,$f^{2}(0)+g^{2}(0)=9$ then prove that there exists some $c \in(-3,3)$ such that $g(c) \circ g^{n}(c) < 0$.
Explanation:
$f(x)$ is differentiable function and $g(x)$ is double differentiable function.
let $f(x) = {{{x^3}} \over 3} + 9,g(x) = {x^2}$
$f'(x) = {x^2}$
So, the function satisfies the criteria $f'(x) = g(x)$
Also, the two function
${f^2}(0) = 0 + 9 = 9$
${g^2}(0) = 0$
$ \Rightarrow {f^2}(0) + {g^2}(0) = 9$
$|f(x)| \le 1$
$ \Rightarrow \left| {{{{x^3}} \over 3} + 9} \right| \le 1$
$ \Rightarrow - 1 \le {{{x^3}} \over 3} + 9 \le 1$
$ \Rightarrow - 10 \le {{{x^3}} \over 3} \le - 8$
$ \Rightarrow - 30 \le {x^3} \le - 24$
$ \Rightarrow {( - 30)^{{1 \over 3}}} \le x \le {( - 24)^{{1 \over 3}}}$
$ \Rightarrow ( - 3.10) \le x \le - 2.88$
Now, $g(c) = {c^2}$
$g'(x) = 2x$
$g''(x) = 2$
It is given $g(c).~g''(c) < 0$
$\Rightarrow 2c^2 < 0$
$\Rightarrow c < 0$
As $x$ lies between $-2.88$ to $-3.10$
So, $c$ lies between $[-3,3]$
prove that ${{y'} \over y} = {1 \over x}\left( {{a \over {a - x}} + {b \over {b - x}} + {c \over {c - x}}} \right)$.
${\left( {\sin y} \right)^{\sin \left( {{\pi \over 2}x} \right)}} + {{\sqrt 3 } \over 2}{\sec ^{ - 1}}\left( {2x} \right) + {2^x}\tan \left( {In\left( {x + 2} \right)} \right) = 0$
that $\left( {{x^2} + 4} \right){\left( {{{dy} \over {dx}}} \right)^2} = {n^2}\left( {{y^2} + 4} \right)$
then show that $\left| {\matrix{ {A\left( x \right)} & {B\left( x \right)} & {C\left( x \right)} \cr {A\left( \alpha \right)} & {B\left( \alpha \right)} & {C\left( \alpha \right)} \cr {A'\left( \alpha \right)} & {B'\left( \alpha \right)} & {C'\left( \alpha \right)} \cr } } \right|$ is
divisible by $f(x)$, where prime denotes the derivatives.
$f''\left( x \right) = - f\left( x \right),$ and $f'\left( x \right) = g\left( x \right),h\left( x \right) = {\left[ {f\left( x \right)} \right]^2} + {\left[ {g\left( x \right)} \right]^2}$
Find $h\left( {10} \right)$ if $h(5)=11$
at $x=1$
and $F\left( x \right) = \left| {\matrix{ {{f_1}\left( x \right)} & {{f_2}\left( x \right)} & {{f_3}\left( x \right)} \cr {{g_1}\left( x \right)} & {{g_2}\left( x \right)} & {{g_3}\left( x \right)} \cr {{h_1}\left( x \right)} & {{h_2}\left( x \right)} & {{h_3}\left( x \right)} \cr } } \right|$ then $F'\left( x \right)$ at $x = a$ is ...........