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Question: If \[\cos x = \dfrac{1}{{\sqrt {1 + {t^2}} }}\] and \[\sin y = \dfrac{t}{{\sqrt {1 + {t^2}} }}\] the...

If cos⁡x=11+t2\cos x = \dfrac{1}{{\sqrt {1 + {t^2}} }} and sin⁡y=t1+t2\sin y = \dfrac{t}{{\sqrt {1 + {t^2}} }} then dydx=\dfrac{{dy}}{{dx}} =
A.11
B.00
C.−1 - 1
D.None of these

Explanation

Solution

Hint : The process of finding a derivative is called differentiation. The reverse process of finding a derivative is called antidifferentiation. The fundamental theorem of calculus relates antidifferentiation with integration. The derivative of a function of a real variable measures the sensitivity to change of the function value with respect to a change in its argument.

Complete step-by-step answer :
In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value with respect to a change in its argument. Derivatives are a fundamental tool of calculus.
We are given cos⁡x=11+t2\cos x = \dfrac{1}{{\sqrt {1 + {t^2}} }}
Put t=tan⁡θt = \tan \theta
Therefore we get x=cos⁡−1[11+tan⁡2θ]x = {\cos ^{ - 1}}\left[ {\dfrac{1}{{\sqrt {1 + {{\tan }^2}\theta } }}} \right]
= {\cos ^{ - 1}}\left[ {\dfrac{1}{{\sqrt {{{\sec }^2}\theta } }}} \right]$$$$ = {\cos ^{ - 1}}\left[ {\cos \theta } \right] = \theta
Therefore x=tan⁡−1tx = {\tan ^{ - 1}}t
On differentiating both the sides we get ,
dxdt=11+t2\dfrac{{dx}}{{dt}} = \dfrac{1}{{1 + {t^2}}}
And we are given sin⁡y=t1+t2\sin y = \dfrac{t}{{\sqrt {1 + {t^2}} }}
Put t=tan⁡θt = \tan \theta
Therefore we get y=sin⁡−1tan⁡θ1+tan⁡2θy = {\sin ^{ - 1}}\dfrac{{\tan \theta }}{{\sqrt {1 + {{\tan }^2}\theta } }}
=sin⁡−1(tan⁡θsec⁡θ)=sin⁡−1(sin⁡θ)=θ= {\sin ^{ - 1}}\left( {\dfrac{{\tan \theta }}{{\sec \theta }}} \right) = {\sin ^{ - 1}}(\sin \theta ) = \theta
Therefore y=tan⁡−1ty = {\tan ^{ - 1}}t
On differentiating both the sides we get ,
dydt=11+t2\dfrac{{dy}}{{dt}} = \dfrac{1}{{1 + {t^2}}}
Hence we get dydx=dydtdxdt=1\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}} = 1
Therefore option (1) is the correct answer.

So, the correct answer is “Option 1”.

Note : The derivative of a function of a single variable at a chosen input value, when it exists, is the slope of the tangent line to the graph of the function at that point. The tangent line is the best linear approximation of the function near that input value. For this reason, the derivative is often described as the "instantaneous rate of change", the ratio of the instantaneous change in the dependent variable to that of the independent variable.