If, f (x, y) = x2 + y3; x = t2 + t3; y = 1 + t3Value of df/dt at t = 1 is:

If, f (x, y) = x2 + y3; x = t2 + t3; y = 1 + t3Value of df/dt at t = 1

1.	If, f (x, y) = x2 + y3; x = t2 + t3; y = 1 + t3Value of df/dt at t = 1 is:
A.	0
B.	2
C.	19
D.	56
Answer» E.

Discussion

No Comment Found

Related MCQs

If \({\rm{y}} = {\sec ^{ - 1}}\left( {\frac{{{\rm{x}} + 1}}{{{\rm{x}} - 1}}} \right) + {\sin ^{ - 1}}\left( {\frac{{{\rm{x}} - 1}}{{{\rm{x}} + 1}}} \right)\), then \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}}\) is equal to
Let \({\rm{f}}\left( {{\rm{x}},{\rm{y}}} \right) = \frac{{{\rm{a}}{{\rm{x}}^2} + {\rm{b}}{{\rm{y}}^2}}}{{{\rm{xy}}}}\), where a and b are constants. If \(\frac{{\partial {\rm{f}}}}{{\partial {\rm{x}}}} = \frac{{\partial {\rm{f}}}}{{\partial {\rm{y}}}}{\rm{\;}}\) at x = 1 and y = 2, then the relation between a and b is
If \(f\left( x \right) = \frac{{{x^3}}}{3} - \frac{{5{x^2}}}{2} + 6x + 7\) increases in the interval T and decreases in the interval S, then which one of the following is correct?
Find the differentiation of \({\cos ^{ - 1}}x\)
If v = yz î + 3zx ĵ + z k̂, then curl v is
Change the order of integration in\(\mathop{\int }_{0}^{a}\mathop{\int }_{y}^{a}\frac{x}{{{x}^{2}}+{{y}^{2}}}dx~dy\)
Each of four particles move along an x-axis. Their coordinates (in meters) as functions of time (in seconds) are given by1) particle 1: x (t) = 3.5 – 2.7 t32) particle 2: x (t) = 3.5 + 2.7 t33) particle 3: x (t) = 3.5 – 2.7 t24) particle 4: x (t) = 3.5 – 3.4t - 2.7 t2Which of these particles have constant acceleration?
Differentiate {-log (log x), x > 1} with respect to x
A function f(x) = 1 - x2 + x3 is defined in the closed interval [-1, 1]. The value of x, in the open interval (-1, 1) for which the mean value theorem is satisfied, is
If \(u = \log \left( {\frac{{{x^2} + {y^2}}}{{x + y}}} \right)\) then \(x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial x}}\) is equal to –

If, f (x, y) = x2 + y3; x = t2 + t3; y = 1 + t3Value of df/dt at t = 1 is:

Discussion

No Comment Found

Related MCQs

Reply to Comment