`(d)/(dx)e^(log) x=` (1) `1,` (2) `0,` (3) `log x,` (4) `(1)/(x)`

The differential coefficient of f((log)_e x) with respect to x , where f(x)=(log)_e x , is (a) x/((log)_e x) (b) 1/x(log)_e x (c) 1/(x(log)_e x) (d) none of these

Differentiate the following functions with respect to x (a) x^(4) + 3x^(2) -2x (b) x^(2) cos x (c) (6x +7)^(4) (d) e^(x) x^(5) (e) ((1 + x))/(e^(x))

(d)/(dx)(e^((1)/(2)log(1+tan^(2)x))) is equal to

d/dx(x^2+e^x+logx+sinx)

d/(dx)[log{e^x((x-2)/(x+2))^(3//4)}] equals (a) (x^2-1)/(x^2-4) (b) 1 (c) (x^2+1)/(x^2-4) (d) e^x(x^2-1)/(x^2-4)

The primitive of the function f(x)=(1-1/(x^2))a^(x+1/x)\ ,\ a >0 is (a) (a^(x+1/x))/((log)_e a) (b) (log)_e adota^(x+1/x) (c) (a^(x+1/x))/x(log)_e a (d) x(a^(x+1/x))/((log)_e a)

Differentiate the following w.r.t. x:(i) e^(-x) (ii) sin (log x), x > 0 (iii) cos^(-1)(e^x) (iv) e^(cosx)

The function f(x)=e^x+x , being differentiable and one-to-one, has a differentiable inverse f^(-1)(x)dot The value of d/(dx)(f^(-1)) at the point f(log2) is (a) 1/(1n2) (b) 1/3 (c) 1/4 (d) none of these

The function f(x)=e^x+x , being differentiable and one-to-one, has a differentiable inverse f^(-1)(x)dot The value of d/(dx)(f^(-1)) at the point f(log2) is 1/(1n2) (b) 1/3 (c) 1/4 (d) none of these

If y(x) is solution of differential equation satisfying (dy)/(dx)+((2x+1)/x)y=e^(-2x), y(1)=1/2e^(-2) then (A) y(log_e2)=log_e2 (B) y(log_e2)=(log_e2)/4 (C) y(x) is decreasing is (0,1) (D) y(x) is decreasing is (1/2,1)