Question

If y=tanA, then dy/dA at A=pi/4 will be ??
a) 1/2
b) 1
c) 2
d) 3/2​

Answer 1

Given :

y= tan A

To find :

dy/dA at A=π/4

Solution

We have ,

$\sf\:y=\tan\:A$

Now Differentiate with respect to A

$\sf\dfrac{dy}{dA}=\dfrac{d(\tan\:A)}{dA}$

$\sf\dfrac{dy}{dA}=\sec^2A$

We have to find dy/dA at A= π/4

$\sf\dfrac{dy}{dA}(A=\dfrac{\pi}{4})=(\sec\dfrac{\pi}{4})^2$

$\sf\dfrac{dy}{dA}(at\:A=\dfrac{\pi}{4})=(\sqrt{2})^2$

$\sf\dfrac{dy}{dA}(at\:A=\dfrac{\pi}{4})=2$

Correct option c) dy/dA( at A=π/4)=2

${\purple{\boxed{\large{\bold{\dfrac{dy}{dA}(at\:A=\dfrac{\pi}{4})=2}}}}}$

$\rule{200}2$

${\underline{\sf{Formula's}}}$

$1)\sf\:\frac{d(x {}^{n} )}{dx} = nx {}^{n - 1}$

$2)\sf\:\frac{d(constant)}{dx} = 0$

$3) \sf\dfrac{d(\cos\:x)}{dx} =-\sin\:x$

$\sf4)\dfrac{d(\tan\:x)}{dx}=\sec^2\:x$

${\red{\boxed{\large{\bold{Composite\: Function (Chain\:Rule)}}}}}$

Let y=f(t) ,t = g(u) and u =m(x) ,then

$\sf\:\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{du} \times \dfrac{du}{dx}$

Answer 2

Explanation:

Given :

y= tan A

To find :

dy/dA at A=π/4

Solution

We have ,

\sf\:y=\tan\:Ay=tanA

Now Differentiate with respect to A

\sf\dfrac{dy}{dA}=\dfrac{d(\tan\:A)}{dA}

dA

dy

=

dA

d(tanA)

\sf\dfrac{dy}{dA}=\sec^2A

dA

dy

=sec

2

A

We have to find dy/dA at A= π/4

\sf\dfrac{dy}{dA}(A=\dfrac{\pi}{4})=(\sec\dfrac{\pi}{4})^2

dA

dy

(A=

4

π

)=(sec

4

π

)

2

\sf\dfrac{dy}{dA}(at\:A=\dfrac{\pi}{4})=(\sqrt{2})^2

dA

dy

(atA=

4

π

)=(

2

)

2

\sf\dfrac{dy}{dA}(at\:A=\dfrac{\pi}{4})=2

dA

dy

(atA=

4

π

)=2

Correct option c) dy/dA( at A=π/4)=2

{\purple{\boxed{\large{\bold{\dfrac{dy}{dA}(at\:A=\dfrac{\pi}{4})=2}}}}}

dA

dy

(atA=

4

π

)=2

\rule{200}2

{\underline{\sf{Formula's}}}

Formula

′

s

1)\sf\:\frac{d(x {}^{n} )}{dx} = nx {}^{n - 1}1)

dx

d(x

n

)

=nx

n−1

2)\sf\:\frac{d(constant)}{dx} = 02)

dx

d(constant)

=0

3) \sf\dfrac{d(\cos\:x)}{dx} =-\sin\:x3)

dx

d(cosx)

=−sinx

\sf4)\dfrac{d(\tan\:x)}{dx}=\sec^2\:x4)

dx

d(tanx)

=sec

2

x

{\red{\boxed{\large{\bold{Composite\: Function (Chain\:Rule)}}}}}

CompositeFunction(ChainRule)

Let y=f(t) ,t = g(u) and u =m(x) ,then

\sf\:\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{du} \times \dfrac{du}{dx}

dx

dy

=

dt

dy

×

du

dt

×

dx

du