Advertisements
Advertisements
Question
Differentiate the function with respect to x.
`sec(tan (sqrtx))`
Advertisements
Solution
Let, y = `sec(tan (sqrtx))`
Differentiating both sides with respect to x,
`dy/dx = d/dx sec [tan (sqrtx)]`
= `sec (tan sqrtx) tan (tan sqrtx) d/dx tan sqrtx`
= `sec (tan sqrtx) tan (tan sqrtx) sec^2 sqrtx d/dx (sqrtx)`
= `sec (tan sqrtx) tan (tan sqrtx) sec^2 sqrtx * 1/2 x^(1/2-1)`
= `sec (tan sqrtx) tan (tan sqrtx) sec^2 sqrtx * 1/(2sqrtx)`
APPEARS IN
RELATED QUESTIONS
Differentiate the function with respect to x.
cos (sin x)
Prove that the function f given by f(x) = |x − 1|, x ∈ R is not differentiable at x = 1.
Differentiate the function with respect to x:
`(cos^(-1) x/2)/sqrt(2x+7)`, −2 < x < 2
Find `dy/dx`, if y = 12 (1 – cos t), x = 10 (t – sin t), `-pi/2 < t < pi/2`.
Does there exist a function which is continuos everywhere but not differentiable at exactly two points? Justify your answer?
If sin y = xsin(a + y) prove that `(dy)/(dx) = sin^2(a + y)/sin a`
If y = tan(x + y), find `("d"y)/("d"x)`
Let f(x)= |cosx|. Then, ______.
Differential coefficient of sec (tan–1x) w.r.t. x is ______.
If u = `sin^-1 ((2x)/(1 + x^2))` and v = `tan^-1 ((2x)/(1 - x^2))`, then `"du"/"dv"` is ______.
|sinx| is a differentiable function for every value of x.
cos |x| is differentiable everywhere.
`sin sqrt(x) + cos^2 sqrt(x)`
`cos(tan sqrt(x + 1))`
sinx2 + sin2x + sin2(x2)
(sin x)cosx
(x + 1)2(x + 2)3(x + 3)4
`tan^-1 (secx + tanx), - pi/2 < x < pi/2`
`sec^-1 (1/(4x^3 - 3x)), 0 < x < 1/sqrt(2)`
`tan^-1 ((3"a"^2x - x^3)/("a"^3 - 3"a"x^2)), (-1)/sqrt(3) < x/"a" < 1/sqrt(3)`
`tan^-1 ((sqrt(1 + x^2) + sqrt(1 - x^2))/(sqrt(1 + x^2) - sqrt(1 - x^2))), -1 < x < 1, x ≠ 0`
If xm . yn = (x + y)m+n, prove that `("d"^2"y")/("dx"^2)` = 0
If k be an integer, then `lim_("x" -> "k") ("x" - ["x"])` ____________.
A function is said to be continuous for x ∈ R, if ____________.
If `y = (x + sqrt(1 + x^2))^n`, then `(1 + x^2) (d^2y)/(dx^2) + x (dy)/(dx)` is
`d/(dx)[sin^-1(xsqrt(1 - x) - sqrt(x)sqrt(1 - x^2))]` is equal to
If `ysqrt(1 - x^2) + xsqrt(1 - y^2)` = 1, then prove that `(dy)/(dx) = - sqrt((1 - y^2)/(1 - x^2))`
If f(x) = `{{:(ax + b; 0 < x ≤ 1),(2x^2 - x; 1 < x < 2):}` is a differentiable function in (0, 2), then find the values of a and b.
If f(x) = `{{:(x^2"," if x ≥ 1),(x"," if x < 1):}`, then show that f is not differentiable at x = 1.
Prove that the greatest integer function defined by f(x) = [x], 0 < x < 3 is not differentiable at x = 1 and x = 2.
